LIBRARY OF CONGRESS. 



Shelf^L.2__ 

UNITED STATES OF AMEPJCA. 



To-day competition in every kind of trade is exceed- 
ingly sharp. Business is not done in the old fashioned 
way, — that most anything is good enough so long as the 
customer don't know of something better. Now-a-days 
your customer is very likely to know a good deal about the 
best goods and about the lowest prices. The live, active, 
wide-awake dealer must know whpre the best goods are 
made and where the most favorable prices can be secured. 
We manufacture stoves. We are not going to say here 
that we make the best stoves in the world and we are not 
going to say that we can sell at the lowest prices. We do 
say that we have particular facilities for making good 
stoves at reasonable prices. 

Now what we want to say to you (YOU ARE A 
STOVE DEALEI?, AREN'T YOU? is that you 
can't afford not to compare all the leading goods made in 
this country before you decide on your lines to carry in 
stock. All we ask is a fair comparison. We ask that 
pretty strongly and we mean it. If our goods and prices 
don't compare favorably, we can't ask anything more 
of you. 

It won't cost you anything (except a postage stamp 
or a postal card) to get our catalogue. We are the Mt. 
Penn Stove Works and our office and foundry are at 
Reading, Pennsylvania. We can't begin to give you in 
this space any idea of our goods (stoves, ranges and fur- 
naces), but we would like to send you our catalogue. 
This is a very modest request, — don't you think so? 



Digitized by tine Internet Arciiive 
in 2011 witii funding from 
The Library of Congress 



http://www.arcliive.org/details/tinsmitlispatternOOIitt 



QQQKING mJlBE BJISY 

..Success the Proof of Me rit!.. ^ 

Over a Quarter Million NEW PRGGCSS (Evaporating) 
Stoves in use in the United States. 




For i8q4 equipped with brass hot air pipes, improved sight 
feed, removable drip valves, with anti-rusting points, cleanable 
reservoir and handsome planished steel asbestos lined oven — the 
best baker made. 

THE STAt\/DARO LIGHTING CO. 

CLEVELAND, OHIO. 



Tte Best nw (or Tiqp. 




PUBLISHED WEEKLY AT 



$2.00 PER YEAR IN ADVANCE. 



THIS JOURNAL IS DEVOTED TO THE 

Stove, TlniHardiare Interests 

And has a practical Tinshop Department in which are published 
tables, receipts, rules and diagrams for 



mm ew 



!!5i*'*'^!*1 



as well as Special Departments devoted to practical papers on Cornice Making-,. 
Hot Air, Steam and Hot Water Heating. 

Answers to questions, etc., which often make a single number worth more 
than the price of a year's subscription. The free want and sales columns have- 
often secured tinners, or tinners have obtained situations, or disposed of tools- 
through them. 

American Artisan Full Size Pattern Sheet 

Comprises full size patterns for over 100 articles of pieced tinware. Price of 
entire set, $1.00, postpaid. 



Address, enclosing currency, express or postoffice money order or draft, 

DANIEL STERN, 

69 Dearborn Street, - - CHICAGO, ILI^ 



JEWEL GBSOLEUE STOYES... 



^ 



WE'RE' AWARDED. 



HIGHEST MEDAL 
AND DIPLOMA 



World's Fair, Chicago, 1 S93. 




SEIND FOR GATTAIaOG. 



George M. Clark & Company 



MAKERS 

149-161 Superior Street, 



Plumbers' and Tingiars' Furnace lo. 2 



Roofing and Plumbing 

Strong Fire; Wind-.' sfc 
Li.s^ht.... 



Our No. 1 

For Tinners' shop use, has 
Double Burner and High 
Tank... Price of either... 
§7.00. 

Discounts to tlie Trade. 



Both the Best in their Line 
Thousands in use... 
We refer to everyone. 



SOLOEBiHG 
FURNACES... 

PBICE, $5.0^. 



Gasoline 
Torches 

For in and out 
door use, etc., etc, 




RINGE 



TCVE CO 



Manufacturers of j^ 

THF. . 

CELEBRATED... 



diriieal'- 



GASOLINE 

GAS 

STOVES. 



THEl 



TiisMiTHS' Patteri Maiual 



Patterns for Tinsmiths' Work. 



\o> 



BY 



/) JOE K. LITTLE, [C. E. 



FOR 



Tinners, Coppersmiths, Plumbers, Zinc Workers, and 
Sheet Metal Workers Generally. 



1894. 

THE AMERICAN ARTISAN PRESS, 

69 Dearborn Street, 

CHICAGO. 




<^^ . \> 






Entered according to Act of Congress in the year 1893, by 

JOSEPH KANE LITTLE, 

In the Office of the Librarian of Congress, at Washington, D. C. 



3 ^ 



.^ 



. . . Preface 



'T^HE purpose of this work is an eminently practical 
-*■ one, as it is designed not so much to furnish a 
batch of isolated patterns in' common use, as to lay 
down general geometrical principles, each one of which, 
when mastered by the reader, will enable him to draw a 
number of different patterns whose principle ot con- 
struction is essentially the same. The sheet metal 
worker who masters the geometrical constructions here- 
with presented, can easily develop the surface of any 
article with much greater ease and rapidity than by^ 
following the various methods in general vogue. 

Concerning the greater utility of the patterns here- 
in shown, I can speak from experience, having served 
a pains-taking apprenticeship in the workshop system 
of setting out patterns before it was my good fortune 
to discover the application of geometrical principles to 
what had been my daily toil. It has been my constant 
aim to make the book a satisfactory one from the stand- 
point of both the mathematician and the workshop, and 
the many enconiums bestowed on it by mechanical en- 



gineers and technical professors as well as by working 
tinners to whom advance sheets have been shown, leads 
me to think that I have succeeded in my aim. 

Each of the problems presented is complete in it- 
self; but although solved independently they follow 
each other in due order. It is my modest hope that this 
work will prove a useful addition to our scanty collec- 
tion of volumes on geometry as applied to sheet metal 
work. 

The Author. 
Philadelphia, Pa., Oct., 1893. 



CONTENTS. 
Book I. 

CHAPTEE L 

TAOE 

Classipicatioit .. « •• •• i 

CHAPTER II. 

Intboductost Pkoblems 3 

Definitions, 3-5 ; Problems on Angles, Lines, Circles, Poly- 
gons, Ovals, Ellipses, and Oblongs, 6-20 ; Meafmrement 
of Angles, 20-3. 

CHAPTER m. 

Aeticles of Equal Taper or IwcLmATioN op Slant .. .. 24 
Right cone defined and described, with problems, 24-7. 

CEAPTER IV. 

Patterns fob Round Articles op Equal Taper or Inclina- 
tion OF Slant .. .. .. .. 28 

Developmeiit of right cone, and problems on same, 28-31 ; 
allowance for lap, seam, and wiring, 32-3 ; frustum (round 
equal-tapering bodies) defined and described, with pro- 
blems, 33-37 ; patterns for round equal-tapering bodies 
(frusta) in one, two, or more pieces, for both email and 
large wort;, 37-45. 



viii CONTENTS. 

CHAPTER V. 

VJiOK 

Eqcal-taperinq Bodies and theie Plavs „ ., .. 46 

Plans and elevations, 46-50 ; plans of equal-tapering bodies ; 
their characteristic features, 50-6 ; problems on the plans 
of round, oblong, and oval equal-tapering bodies, 56-65 

CHAPTEE VI. 
Patterns foe Flat- paced Equal-tapeking Bodies .. ., 66 

Definition of right pyramid, and development, 66-7 ; pattern 
for right pyramid, 68-9 ; frustum of right pyramid defined 
and described, 69-71 ; patterns for frusta of right pyramids 
(hoppers, hoods, &c.) for both small and large work, 71-7 ; 
baking-pan pattern in one or more pieces, 77-83. 

CHAPTER VII. 

Patterns for Equal-tapebing Bodies of Flat and Curved 

Surfaces combined .. .. .. .. .. .. 84 

Pattern for oblong body with flat sides and semicircular ends, 
84r-90; for oblong body with round comers, 90-6; for 
oval body, 96-104. 

The patterns for each of these bodies are given in one, two, 
and four pieces, as well as for both small and large work. 



Book II. 

CHAPTER I. 

Patterns for Bound Articles of Unequal Taper ob Incli- 
nation op Slant .. .. .. .. .. .. 105 

Oblique cone, definition, description, and development, 
105-111 ; oblique cone frustum (round unequal-tapering 
body) and oblique cylinder, 111-3; patterns for roimd 
unequal-tapering bodies (oblique cone frusta), also 
oblique cylinder — work small or large, 113-23 



CONTENTS. IX 

CHAPTEE 11. 

PAGB 

TTNEQtrAL-TAPERING BODIES AND THEIR PLANS .. .. .. 124 

Plans of uuequal-tapering bodies and their characteristic 
features, 124-9 ; problems on plans of oblong, oval, and 
other unequal-tapering bodies, 130-3 ; plans of Oxford 
hip-bath, Athenian hip-bath, sitz bath and oblong taper 
bath. 134-42 

CHAPTER III. 
rATTERys FOR Flat-faced Unequal-tapering Bodies .. ., 143 

Oblique pyramids and their frusta, 143-5 ; pattern for oblique 
pyramid, 145-8 ; patterns for frusta of oblique pyramids 
(flat-faced unequal- tapering bodies) — work small or large, 
148-54; pattern for a hood which is not a frustum of 
oblique pyramid, 154-6. 



CHAPTER IV. 

Patterns for IJNEQUAL-TAPERiNa Bodies of Flat and Curved 

Surfaces combined .. ,. .. ., .. .. 157 

Pattern for equal-end bath (unequal-tapering body having 
flat sides and semicircular ends), 157-68 ; for oval bath, 
(oval unequal-tapering body), 168-83; for tea-bottle top 
(unequal-tapering body having round top, and oblong 
bottom with semicircular ends), 183-91 ; for oval-canister 
top (unequal-tapering body having round top and oval 
bottom), 192-205 ; for unequal-tapering body having 
round top and oblong bottom with round comers, 205-16. 
( The patterns for each of these bodies are given in one, two, or 
four pieces, as well as for both small and large work.^ 
Pattern for Oxford hip-bath, 216-30 (two methods) ; for 
oblong taper bath, 230-6 (two methods). 



ETAL-PLATE WORK. 



BOOK I. 

CHAPTEE I. 

Classification. 



(1.) Notwithstanding the introduction of macliinery and 
the division of labour in the variotss "branches of metal- 
plate work, there is as, great a demand for good metal-plate 
workers as ever, if not indeed a greater demand than 
formerly, while the opportunities for training such men are 
becoming fewer. An important part of the technical edu- 
cation of those connected with sheet-metal work is a know- 
ledge of the eetting-out of patterns. Such knowledge,, 
requisite always by reason of the variety of shapes that 
are met with in articles made of sheet-metal, is nowadays 
especially needful ; in that the number of articles made of 
sheet metal, through Ihe "revival of art metal-work, the 
general advance of science, and the introduction of new 
designs (which in many cases have been very successful), 
ia articles of domestic use, has considerably increased. It 
is with the eetting-out of patterns that this volume princi- 
pally deals. To practical men, the advantages in saving 
of time and material, of having correct patterns to work 
from, are obvious. Whilst, however, the method of treat- 
ment here of the subject will be essentially practical, an 



THE TINSMITHS- PATTERN MANUAL. 



HiiKjunt of tlieory' sufficiont f6r a thorough comprehension 
of the rules given will he introduced, a knowledge of rules 
without principles being mere ' rule O'f thumb,' and not true 
technical education. 

(2.) Starting in the following pages with some introduc- 
tory problems and other matter, we shall proceed from these 
to the articles for which patterns are required by sheet-metal 
Avorkers and which may be thus conveniently classed and 
xnbdivided : \ 

ia. Of round surfaces. 
6. Of plane or flat 
surfaces, 
divisions. | c. Of curved and 
I plane surface 

\ combined. 



Class T, — Patterns for Ar- 
ticles of e(lual taper 
or inclination (pails, 
oval teapots, gravy 
strainers, &c.). 

Class 11. — Patterns for Ar- 
ticles of unequal taper 
or inclination (baths, 
hoppers, canister - 
tops, &c.). 



Sub- 
divisions. 



o. Of round surfaces. 
6. Of plane or flat 

surfaces, 
c. Of . curved and 

plane surface 

combined. 



Class III. — Patterns for Miscellaneous Articles (elbows, and' 
articles of compound bent surface, as vases, aquarium 
stands, mouldings, &c.). 

All these articles will be found dealt with in their several 
planes. 

We shall conclude with a few technical details in respect 
of the metals that metal-plate workers mostly make use of. 

(3.) The setting out of patterns in sheet-metal work 
belongs to that department of solid geometry known as 
" Development of Surfaces," which may be said to be the 
Rpreading or laying ' out without rupture the surfaces of 
solids in the plane or flat, the plane now being sheet metal. 



llii-: TlN'SMITliS' PATTERN MANIJA1-. 



CHAPTEK II. 

Introdtjctouy Problems ; with Applications. 

Definitions. 

Straiglit Line. — A straight line is the shortest distance 
between two points. 

Note. — If not otherwise stated, lines are always supposed to be straiglit. 

Angle. — An angle is the inclination of two lines, which 
meet, one to another. The lines A B, C B in Fig. 1 which 
are inclined to each other, and meet in B, are said to form 
an angle with one another. To express an angle, the letters 
which denote the two lines forming the angle are employed, 
the letter at • the angular point being placed in the middle j 
thus, in Fig. 1, we speak of the angle ABC. 



Fig. 1. 



Fig. 2. 
A 



Fis. 3. 





Perpendicular. Biglit Angles. — If a straight line, A B 
(Fig. 2), meets or stands on another straight line, G D, so that 
the adjacent angles (or angles on either side of A B) A B I>, 
A B C, are equal, then the line A B is said to be perpendicular 
to, or at right angles with (' square with ') D G, and each of 



the angles is a right angle. 



4 THE TINSMITHS' PATTERN MANUAL. 

Parallel Lines. — Parallel lines are lines wHich, if produced 
ever so far both ways, do not meet. 

Triangle. — -A figure bounded by three lines is called a 
triangle. 

A triangle of whicli one of the angles is a right angle is 
called a right-angled triangle (Fig. 3) ; and the side which 
joins the two sides containing the right angle is called the 
hypothenuse (or hypotenuse). If all the sides of a triangle 
are equal, the triangle is Equilateral, If it has two sides 
equal, the triangle is Isosceles. If the sides are all unequal, 
the triangle is Scalene. 

Polygon. — A figure having more than four sides is called 
a. polygon. Polygons are of two classes, regular and irregular. 

Irregular Polygons have their sides and angles unequal. 

Regular Polygons have all their sides and angles equal, 
and possess the property (an important one for us) that they 
can always be inscribed in circles ; in other woi ds, a circle 
can always be drawn through the angular points of a regular 
polygon (Figs. 12 and 13). 

Special names are given to regular polygons, according to 
the number of sides they tpossess ; thus, a polygon of five 
sides is a pentagon ; of six sides, a hexagon ; of seven, a 
heptagon; of eight, an octagon; and so on. 

Quadrilaterals. — All figures bounded by four lines are 



Fig. 4a. 



Fig. 46. 



called quadrilaterals. The moat important of these are tiie 
square and oblong or rectangle. In a square (Fig. 4rt) tl.e 
sides are all equal and the angles all right angles, and con- 
sequently equal. An oblong or rectangle has all its anglos 
right^ angles, but only its^oppusity sides are_e(iu£il._ (Fig. dl*.) 



TUE TINSMITHS' PATTERN MANUAL. f) 

Circle. — A circle is a figure bounded by a curved line such 
that all oints in the line are at an equal distance from a certain 
point within the figure, which point is called the centre. 

The bounding line of a circle is called its circumference. A 
part only of the circumference, no matter how large or small, 
is called an arc. An arc containing a quarter of the circum- 
ference is a quadrant. An arc containing half the circumfer- 
ence is a semicircle. A line drawn from the centre to any 
point in the circumference is a radius (plural, radii). The 
lino joining the extremities of any arc is a chord. A chord 
that passes through the centre is a diarneter. 

A line drawn from the centre of, and perpendicular to, any 
chord that is not a diameter of a circle, will pass through its 
centre. 

In practice a circle, or arc, is ' described ' from a chosen, or 
given, centre, and with a chosen, or given, radius. 

If two circles have a common centre, their circumferences 
are always the same distance apart. 




In ¥ig. 5. 

O is the centre. 

AD (the curve) is an arc. 

A B (or B C) is a quadrant. 

A C B is a semicircle. 

O A (or B, or B C) is a radius. 

A D (the straight line) is a chord, 

A is a diameter. 



6 THE TiNSMiTllS' PATTERN M\XUAL. 

PROBLEM I. 

To draw an angle equal to a given angle. 

Case I. — Where the * given ' angle is given by a drawing. 

This problem, though simple, is often very useful in practice, 
especially for elbows, where the angle (technically called 
' rake ' or ' bevil ') is marked on paper, and has to be copied. 

Fig. 6. 





Let ABC (Fig. 6) be the given angle. With B as centre 
and radius of any convenient length, describe an arc cutting 
B A, B C (which may be of any length, see Def.) in points A 
and C. Draw any line D E, and with D as centre and same 
radius as before, describe an arc cutting D E in E. With E as 
centre and the straight line distance from A to C as radius, 
describe an arc intersecting in F the arc just drawn. From 
D draw a line through F ; then the angle F D E will be equal 
to the given angle ABC. 

Case II. — Where the given angle is an angle in already 
existing fixed work. 

The angle to which an equal angle has to be drawn, may 
be an angle existing in already fixed work, fixed piping for 
instance ; or in brickwork, when, suppose, a cistern may 
have to be made to fit in an angle between two walls. In 
such cases a method often used in practice is to open a two- 
fold rule in the angle which is to be copied. The rule is then 
laid down on the working surface, whatever it may be (paper, 
board, &c.), on which the work of drawing an angle eqxial to 
the existing angle has to be carried out, and lines are drawn 
on that surface, along eithor the outer ox inner edges of the 



THE TINSMITHS' PATTERN MANUAL. 7 

rule. The rule being tnen removed, tne lines are proauced ; 
meeting, they give the angle required. 

Case III. — Where the given angle is that of fixed work, 
and the method of Case IL is inapplicable. 

With existing fixed work, the method of Case II. is not 
always practicable. A corner may be so filled that a rule 
cannot be appliedi The method to be now employed is as 
follows. Draw lines on the fixed work, say- piping, each way 
from the angle ; and on each line, from the angle, set off any 
the sa.me distance, say 6 in., and measure the distance between 
the free ends of the 6-in. lengths. That is, if A C, A B (Fig. 7) 

Fig. 7. 



represent the lines drawn on the piping, measure the distance 
between B and 0. Now on the workin g^surface on which 
the drawing is to be made, draw any 1 me D E, 6 in. long; 
and with D as centre and radius D E, describe an arc. Next, 
with E as centre, and the distance just measured between B 
and as radius, describe an arc cutting the former arc in F. 
Join F D ; then the angle F D E will be equal to the angle 
of the piping. 

Note. — When poiata are 'joined,* it is always by straight lines. 

PROBLEM IL 

To divide a line into any nuinhcr of equal parts. 
Let A B (Fig. 8) be the given line. From one of its extre- 
mities, say A, draw a line A 3 at any angle to AB, and en it, 
from the angular, point, mark off as many parts, — of any con- 



8 THK TIN.-s^riTHS' I'ATTKRN MANUAL. 

verrient length, but all eqnal to each other, — as A B is to he 
divided into. Say that A B is to he divided into three equal 
jjarts, and that the equal lengths marked oif on A3 are A to 

Fig. 8. 



A C D B 

1, 1 to 2, and 2 to 3. Then join point 3 to the B extremity 
of A B, and through the other points of division, here 1 and 2, 
draw lines parallel to 3 B, cutting A B in C and D. Then 
A B is divided as required. 

PROBLEM III. 

To hiseci (divide a line into two equal parts) a given line. 

Let A B (Fig. 9) be the given line. With A as centre, and 
any radius greater than half its length, describe an indefinite 
arc; and with B as centre and satoe radius, describe an arc 
intersecting the former arc in points P and Q. Draw a line 
through P and Q , this will bisect A B. 




Note. — ^It is quits as easy to bisect A B by Problem II. ; but tbe method 
shown gives, in P Q, not only a line bisecting A B, but a line oerpendicular 
to A B, This mubt be particularly remembered, 



I'HE T[NSM[THS- PATTERN MANUAL. 9 

PEOBLEM IV. 
To find the centre of a given circle. 

Let ABC (Fig. 10) be the given circle. Take any three 
points A, B, C, in its circumference. Join A B, B C ; then 
A B, B C, are chords (see Def.) of the circle ABC. Bisect 
AB, B C ; the point of intersection, 0, of the bisecting lines 
is the centre required. 

Fig. 10. 




PEOBLEM V. 

To describe a circle which shall pass through any three given 
points that are not in the same straight line. 

Let A, B, C (Fig. 10) be the three given points. Join 
A B, B C. Now the circle to be described vsrill not be a circle 
through A, B, C, unless A B, B C, are chords of it. Let us 
therefore assume them such, and so treating them, find (by 
Problem IV.) O the centre of that circle. With as centre, 
and the distance from O to A as radius, describe a circle; it 
will pass- also through B and C, as required. 



PEOBLEM VI. 

Given an arc of a circle, to complete the circle of which it is a 

portion. 

Let A C (Fig. 10) be the given arc ; take any three points 
in it as A, B, C ; join A B, B C. Bisect A B, B C by lines 



16 TUB TINSMITHS' PATTERN MANUAL, 

intersecting in 0. With O as centre, and to A or to any 
puint in the arc, as radius the circle can be completed. 

PEOBLEM VII. 
To find icheiher a given curve ia an are of a circle. 

Choose any three points on the given curve, and by 
Problem V describe a circle passing through them. If the 
circle coincides with the given curve, the curve is an arc. 

PEOBLEM VIII. 

To bisect a given angle. 

Let ABO (Fig. 11) be the given angle. With B as 
centre and ^ny convenient radius describe an arc cutting 
A B, B in D and E. With D and E as centres and any 
co£V3uieut distance, greater than half the length of the aro 




I) E as radius describe arcs intersecting in F. Join F to B j 
then F B bisects the given angle. 

PEOBLEM IX. 

In a given circle, to inscribe a regular polygon of any giasin 
number of sides. 
Divide (Problem II.) the diameter A C of the given circle 
(Fig. 12) into as many equal parts as the figure is to bava 



TIIK TINSMITHS' PATTERN MANUAL. ll 

sides, here say five. With A and C as centres, and C A as 
radius, describe arcs intersecting in P. Through P and the 
second point of division of the diameter draw a line P B 




cutting the circumference in B ; join B A, then B A will be 
one side of the required figure. Mark off the length B A 
from A round the circumference until a marking off reaches 
B. Then, beginning at point A, join each point in the 
circumference to the next following ; this will complete the 
polygon. 

Note. — By this problem a circumference, and therefore also one-half of 
it (semicircle), one-third of it, one-fourth of it (quadiant), aud so on, can 
be divided into any number of equal parts. 

PEOBLEM X. 

To describe any regular polygon, the length of one side being 

given. 
Let A B (Fig. 13) be the given side of, say, a hexagon. 
With either end, here B, as centre and the length of the 
given side as radius, describe an arc. Produce A B to cut 
the arc in X. Divide the semicircle thus formed into as 
many equal parts (Problem IX,, Note) as the figure is to have 
sides (six), and join B fo the second division point of the 
semicircle couniing from X. This line will be another side 
of the required polygon. Having now three points, A, B, 



12 TIIK TINSMITHS' PATTERN MANUAL. 

and the second division point from X, draw a circle tKrough 
them (Problem V.), and, as a regular polygon oan always be 
inscribed in a cii'olo (cee Def.), mark off the length B A round 

Fig. 13. 




the ciroumferenoe from A until at the last marking-off, the 
free extremity of the second side (the side foxind) of the 
polygon is reached, then, beginning at A, join each point in 
the oirotimferenoe to the next following ; this will oomploto 
the polygon (hexagon). 

PROBLEM XL 

To find (he length of the cirmmference of a circle, the diameter 

being given. 

Divide the given diameter A B (Fig. li) into seven equal 

parts (Problem II.). Then three times A B, with C B, on© of 

the seven parts of A B, added, that is with one-seventh of 

FiQ. U. 




B 



A B added, will be the required length of the ciroumferenoe. 
The semicircle of the figure is superfluous, but may help to 
make the problem more clearly understood. 



THE TINSMITH.'?' PATTERN MANUAh 



13 



PROBLEM XII. 
To draw an oval, its length and width being given. 
Draw two lines A B, C D (tho axes of the oval), perpendi- 
cular to one another (Fig. 15), and intersecting in 0. 

Fig. 15. 



Make 




A and B each equal to half the length, and C and D 
each equal to, half the width of the oval. From A mark off 
A E equal to C D the width of the ovalj'and divide E B into 
three equal parts. With 0* as centre and radius equal to 
two of the parts, as E 2, describe arcs cutting A B in points 
Q and Q'. With Q and Q' as centres and QQ' as radius 
describe arcs intersccHng C D in points P and P'. Join. 
P Q, P Q', P' Q and- P' Q' ; in these lines produced the end 
and side curves must meet. With Q and Q' as centres and 
Q A as radius, describe the end curves, and with P and P' as 
centres and radius P D, describe the side curves ; this will 
complete the oval. 

Note. — Unless care is taken, it may be fmind that the end and siue 
curves will not meet accurately, and even with care tlxis may sometimes 
occur. It is best if great accuracy be required in tlie length, to draw the 
end curves first, and tlieu draw side curves to meet them; or, if the width 
is most important, to draw the side curves first. The centres (P and P') 
for the side curves come inside or outside the curves, according as the oval 
is broad or narrow. This figure is sometimes erroneously called an 
ellipse. It is, however, a good approximation to one, and for most purposes 
where aa elliptical article bus to be made, is very convenient. 



14 



TliE TIKSMITHS' PATTERN MANUAL. 



PEOBLEM XIIi: 

To draw an egg-shaped oval, having the length and width given. 

Make A B (Fig. 16) equal to the length of the oval, and 
from A set off A ec^ual to half its width. Through draw 



Fig. 16. 




an indefinite line Q Q' perpendicular to A B, and with as 
centre and A as radius describe the eemicircle CAD. 
Join D B ; and from D draw D E perpendicular to Q Q' and 
e<^ual to D. Also from E draw E G- parallel to Q Q' and 



n\E TINSMITHS PATTERN MAN'UAli. 15 

intersecting D B in Gr, and from G draw G F parallel to D E 
and intersecting Q Q' in F. From B set off B P equal to D F, 
and join P F. Bisect F P and througli the point of bisection 
draw a line cutting Q Q' in Q. Join Q P and produce it 
indefinitely, and with Q as centre .and Q D as radius 
descriho an arc meeting Q F produced in H, Make Q' 
equal to Q, and join Q' P and produce it indefinitely. 
With Q' as centre, and Q' (equal to Q D) as radius, 
describe an arc meeting Q' P produced in H'. And with P 
as centre and P B as radius describe an arc to meet the arcs 
D H and C H' in H and H' ; and to completo the egg-shaped 
oval. 

PEOBLEM XIY. 

To describe an ellipse. 

Before working this as a problem in geometry, let us draw 
an ellipse non-geomatrically and get at some sort of a defini- 
tion. This done,, we Mdll solve the problem geometrically, 
and follow that with a second mechanical method of de- 
scribing the curve. 

METHOD I.— Mechanical. 

A. Irrespective of dimensions. — On a piece of cardboard or 
smooth-faced wood, mark off any two points F, F' (Fig. 17) 
and fix pins securely in those points. Then take a piece of 
thin string or silk, and tie the ends together so as to form a 
loop ; of such size as will pass quite easily over the pins. 
Kes:t, placQ the point of a pencil in the string, and take up 
the slack so that the string, pushed close against the wood, 
shall form a triangle, as say, F D F', the pencil point being 
at D. Then, keeping the pencil upright, and always in the 
string, and the string taut, move the pencil along from left 
to right say, so that it shall make a continuous mark. Let 
us trace the course of the mark. Starting from D, the 
pencil, constrained always by the string, moves from D to F, 



16 



THE TINSMITHS' PATTERN MANUAT.. 



then on to B, T', C, P^, P^, A, P*, and D again, describing a 

curve- which returns into itself ;* this curve is an ellipse. 

Having d^a^vn the ellipse, let us remove the string and 
pins, draw a line from F to F', and produce it both ways to 
terminate in the curve, as at B and A. Then A B is the 
major axis of the ellipse, and F, F' are its foci The mid- 
point of A B is the centre of the ellipse. Any line through 
the centre and terminating botli ways in the ellipse is a 
diameter. The major axis is the longost diameter, and is 
commonly called" the length of the ellipse. The diameter 
through the centre. at right angles to the major axis is the 
shortest diameter, or minor axis, or width of the ellipse. 

Referring to the Fig. : — 

ADPBC is an ellipse. 

F, F' are its foci (singular^/ocus), 

A B is the major axis. 

C D is the minor axis. 
O is the centre. 




We notice with the string and pencil that ^vhen the 
pencil i)oint reaches P, the triangle formed by the string is 
F P F' ; when it reaches P', the triangle is F P' F' ; when it 
reaches P- the triangle is F P^ F' ; and when P^ is reached, 
it id F P^F'. Looking at these triangles, it is obvious that 



THE TINSMITHS' PATTERN MANUAIi. 17 

F F' is ono side of eacH of them ; from which it follows, 
seeing ths:t the loop of string is always of one. length, that 
the siira of the other two sides of any of the triangles is 
equal to the sum of the other two sides of any other of them ; 
that is to say, F D. added to DF' is equal to F P added to 
P F', is equal to P P' added to F F', and go on. 

Which leads us to the following definition. 
Definition, 

EUi})se. — The ellipse is a closed curve (that is, a curve 
returning into itself), such that the sum of the distances of 
any point in the curve from certain two joints (foci), inside 
the curve is always the same. 

B. Length and loidtTi given. — ■ Knowing now what an 
ellipse is, we can work to dimensions. Those usually given 
are the length (major axis), and width (minor axis). Draw 
A B, C D (Fig. 17), the given axes, and with either ex- 
tremity, C or D, of the minor axis as centre, and half A B, 
the major axis as radius, describe an arc cutting AB in F 
and F'. Fix pins securely in F, F' and D (or C). ' Then, 
having tied a piece of thin string or silk firmly round the 
three pins, remove the pin at D (or C) ; put, in place of it, 
a pencil point in the string j and proceed to mark out the 
ellij)se as above explained. 

METHOD II. — Geometrical. — The Solution of the 
Problem. Length and Width- gi\^en. 

Draw A B, C D (Fig. 17), the major and minor axes. With 
C or D as centre, and half the ijiajor axis, B say, As radius, 
describe arcs cutting A B in F and F'. On A B^and between 
O and F', mark points — any number and anywhere, except 
that it is advisable to mark the points closer- to each other 
as they approach F'. Let the points here be 1^ 2, and 3. 
With F and F' as centres and. A 2, B 2 as radii respectively, 
describe arcs intersecting in P; with same centres and A3, 
B 3 as radii respcL-tively, describe arcs intersecting in P'. 
With F' and F as centres and A3, B 3 as radii respectively, 





18 



T1!K TIN.'.MITIIS' PATTKUN MANUAL. 



describe arcs intersecting in P^. With same centres and 
A 2, B 2 aa radii respectively, describe arcs intersecting in 
P*. .Similarly obtain P^. We have thus nine points, D, P» 
B, P', Cj P^, P^j A and P*, through wh"° jh an even curve may 
1)6 drawn which will be the ellipse required. A greater 
number of points through which to draw the ellipse may of 
course be obtained by taking more points between and F', 
and proceeding as explained. 

METHOD III.— SIechanical. — Length and Width '^ 

GIVEN. 

As it is not always possible to proceed as described 
at end of Method L, for pins cannot always be fixed in the 
material to be drawn upon, we now give a second mecha- 
nical method. Having drawn (Fig. 186) AB, CD, the 



Fia. 18a. 
JTnP 



Fig. 18&. 



jfU 




given axes, then, on a strip of card or stiff paper X X (Fig. 
18a), mark off from one end P, a distance PF equal to half 
the major axis (length), and a distance .P E equal to half the 
minor axis (width). Place the strip on the axes in such 
a position that the point B is on the major axis, and the 



TIIR TlXi^MITtlS' PATTERN ^MANUAL. 



19 



point F on the minor, and mark a point against the point P. 
Now shift X X to a position in which E is closer to B, and 
F closer to C, and again mark a point against P. Proceed 
similarly to mark other points, and finally draw an even 
curve through all the points that have been obtained. 



The following problems deal with sha,pes often required 
It/ the metal-plate worTier^ and will give him an idea of how 
to adapt to his requirements the problems that precede. The 
explanation of the measurement of angles that concludes the 
chapter will further assid him in his worTs. 

PROBLEM XV. 

To draw an oblong with round corners. 

Draw two indefinite lines A B, CD (Fig. 19) perpen- 
dicular to one another and intersecting in 0,. Make OA 

Fra 19. 




and 0. B each equal to half the given length; and OC and 
O D each equal to half the given width. Through and D 

n 2 



25 



tUE TlXSMITllS- PATTERN MANUAL 



draw lines parallel to A B, and throngh A and B draw lines 
jiai'allel to C D. We now liave a rectangle or oLlong, and 
re(|mre to roimd tlie corners, which aie quadrants. Mark 
oft" from E along E D and E A equal distances E (t and E F 
according to the size of corner required. With F and G as 
centres and E F or E G as radius, describe arcs intersecting 
in O'. With O' as centre and same radius descril>e the 
corner F G. The remaining corners can be drawn in similar 
manner. 



ITtOBLEM X¥L 

To draw a figure Tiacing straifjJit sides and semicircular ends 
{ohlong with semicircular ends). 

Draw a line AB (Fig. 20) equal to the given length, 
make AO and BO' each equal to half the given widLh. 

Fio. 20. 



I) 




f; 


.( 




"\ 


n 





€ 


y 



F 



G 



Through and 0' draw indefinite lines perpendicular to 
A B; with O and 0' as centres and OA as radius describe 
arcs cutting the perpendiculars through O and O' in D F 
and GE. Join D E, GF; this will complete the figure 
required. 

ANGLES AND THEIR MEASUREMENT. 

The right angle BOG (Fig. 5) subtends the quadrant B C. 
If we divide that quadrant into 90 parts and call the partb 



THE TIK.SMITllS- PATTERN MANUAL. I'l 

degrees, tlaen a rigM angle subtendiS or contains 90 degrees 
(written 9(J^), or as usually espressed, is an angle of 90 
degi-ees, the degree being the unit of measurement. If each 
division point of the quadrant is joined to O, the right 
anglo) is divided into 90 angles, each of which subtends c* is 
an angle of 1 degree. That is to say, an angle is measured 
by the number of degrees that it contains. Suppose the 
quadrant B A is divided as was B C, then BOA also is 
an angle of 90 degrees. If the division is continued round 
the semicircle ADC. this will contain 180 degrees, and the 
whole circumference has been divided into 360 degrees. As 
an angle of 90, which is a fourth part of 360 degrees, subtends 
a quadrant or fourth part of tlie cii'cumference of the circle, 
so an angle of 60, which is a sixth part of 360 degrees, 
subtends a sixth part of the circumference, and similarly an 
angle of 30 degrees subtends a twelfth part, an angle of 45 
an eighth part, and so on. And this angular measurement 
is quit$ independent of the dimensions of the circle; the 
quadrant ahvays subtends a right angle; th.e 60 degrees 
angle always subtends an arc of one sizth of the circum- 
ference ; and the like with other angles. From our defini- 
tion p. 5 we have it tliat a chord is the line joining the 
extremities of any arc. The chord of a sixth part of the 
circumference of any circle, we have now to add, is equal to 
the radius of that circle. This being the case, and as an 
angle of 60 degrees subtends the sixth part of the circumfer- 
ence of a circle, it follows that an angle of 60° subtends a 
chord equal to the radius. 

SCALE OF CHORDS. 

Construction. — We have now the knowledge requisite fcr 
setting out a scale of chords, by wliich angles may be drawn 
and measured. 

On any line B (Fig. 21) describe a semicircle A E, and 
fi-om its centre C draw C A j)erpendicular to B. Divide 
A into nine equal parts. Then, as A, being a qxiadrantj 



^ 



THE TINSP4IT1IS' PATTERN MANUAL. 



^atains 90^, each of the nme divisioBS will contain 10''. The 
points of division, from O, of the quadrant, are marked 10, 20, 
30, &0.J up to 90 at A. With O as centre, describe arcs from 
each of these division points, cutting the line O B. Note that 
the arc from point 60 cuts O B in C, the centre of the semi- 
circle ; the chord from O to 60 (not drawn in the Fig.), that 



Fia. 22. 




O 10 20 30 <J0 So 60 70 60 90 



is, the chord of one-sixth of the circumference of the circle 
whose centre is C, being equal to the radius of that circle-' 
Draw a line O E parallel to O B, and from O let fall O 
perpendicular to OK Also from each of the points where 
the arcs cut B let fall perpendiculars to B and number 
these consecutively to correspond with the numbers on the 
quadrant O A. The scale is now complete. 

Mova touse. It is used in this way. Suppose from a point 
A in any line A B (Fig. 22) we have to draw a line at an 
angle of 30° with it. Then with A as centre and the distance 
from O to 60 on OS (Fig. 21) as radius, describe an arc C D 
cutting A B in G, And with G as centre and the distance 
from O to 30 on E (Fig. 21) (the angle to be drawn is to be 
of 30°) as radius describe an arc intersecting arc D in D. 
Join D A, then DAG will be the required angle of 30°. 
Similarly with angles of other dimensions. 

In taking the lengths of arcs, wo really take the length of 
their chords, and it, is these lengths that (Fig. 21) we have 



tllE TINSMITHS' PATTERN MANUAL. 23 

Bet off along E. The angle (Fig. 21) C F (the point F is 
the point 60) being an angle of 60° subtends a chord equal to 
the radius ; therefore in O to 60 we have the radius C O. In 
the example (Fig. 22), the distance C D (0 to 30) is the chord 
of 30° ; and it is clear that we must set this off on an arc C D 
of a circle of the same size as that employed in the construc- 
tion of the scale, and this we do by making A C equal O to 60 
on the same scale. 

When a scale of chords has been constructed as explained, 
the semicircle may be cut away, and we thus get a scale 
convenient for shop use in the form of a rule. 



24 THE TINSMITHS PATTERN MANUAL 

CHAPTER III. 

Patterits for Articles of Equal Taper or Incldtation. 

(CLASS I.) 

(4.) It is necessary here at once to remark that ©rdinary 
worlisLop parlance speaks of' slant/ — not as meaning an angle, 
bnt a length ; not as refeiring to the angle of inclination of a 
tapering body, but to the length of its slanting portion. It 
is in this sense that we shall nse the word, and shall employ 
the word ' taper ' or tiie term ' inclination of slant ' when 
meaning an angle. 

(5.) In order that the rales for the setting out of patterns 
for articles of equal taper or inclinntion may be better under- 
s|cod and remembered, it is advisable to consider the principles 
on which the rules are based, as a knowledge of principles 
will often enable a workman himself to find rules for the 
setting out of ])attem8 for odd wurk. The basis of the whole 
of the articles in this Class is the right ci>ne. It is nee ssary, 
therefore, to define the ri^ht cone and exolain some of its 
properties. 

Det-inition. 

(6.) Bight Coup. — A right cone Is a solid fi.^ure gcnoratsd 
or formed by the' revolution of a rfght-arglcd triangle about 
one of the sides containing the right angle. The side 
about which the triangle revolves is the axis of the cone ; 
the other side containing the right angle being its radius. 
The point of the cone is its apex; the circular end its hase. 
The hypotenuse of the triangle is the slant of the cone. 
From the method of .formation of the right cone, it follows 
that the axis is perpendicular to the base. The height of 
the cone is the length of its axis 

(7.) Referring to Fig. la, OBE represents a coae gene- 



THE TINSMITHS' PATTERN MANUAL. 



25 



rated or formed by the revolution of the right-angled tri- 
angle A B (Fig. 16) about one of its sides containing the 
right angle, hero the side A, Similarly the cone O D F, 




Fig. 2a, is formed by the revolution of C D (Fig. 26) about 
its side . O C. As will be seen from the figs., O A, O C are 
respectively the axes of the cones B E, O D F, as also their 
heights. Their bases are respectively B G E H, D K F L, 



F'G. 25. 




and the radii of the bases are A B and C D. The slants of 
the cones are O B and O D, the apes in either being the point 
O. Other lines will be seen in figs,, namely, those repre- 
senting the revolving triangle in its motion of generating 



26 



TtiK TJNS-MrTUS' PAT'I EKN MAXl'AL. 



the cone. The sides of tliese triangles that start from the 
apex and terminate in the base are all equal, it must he home 
in mind ; and each of them is the slant of the cone. Likewise 
their sides that terminate in A are all equal, and each shows 
a radius of the base of the cone. How these particulars of 
the relations to one another of the several parts of the right 
cone apply in the setting-out of patterns will be seen in the 
problems that follow. 

PEOBLEM I. 

To find the height of a cone, the slant and diamsier of the hass 
being given. 

m 

Draw any two lines A, B A (Figs. 3 and 4) perpen- 
dicular to each other and intersecting in A. On either line 



Fig. 3. 



Fig. 4. 




_.4 iLouoA-ue of£as& 3 




mark off from A half the diameter of the base, in other 
words, the radius of the base, as A B. With B as centre, and 
radius equal to the slant, describe an arc cutting A in 0. 
Then A is the height of the cone. 



TIIK TTNSMITHS' PATTERN MANUAL. 27 



rEOBT.EM II. 

TofindtTie slant of a cone, the height and diameter of the base 
being given. 

Draw any two lines O A. B A (Figs. 3 and 4") perpendicular 
to each other and intersecting in A. On either line mark off 
from A half the diameter of the base (radins of the base), as 
A Bj and make A O on the other line equal to the height of 
the cone ; join B. Then B is the rec[nired slant 



THE TINSMITHS' PATTERN MANUAL. 



CHAPTEK lY. 

Patteens FOR Round Articles of Equal Taper or 
Inclination of Slant. 

(Class I. Subdivision a.) 

(8.) If a cone has its inclined or slanting surface painted 
say, white, and be rolled while wet on a plane so that every 
portion of the surface in succession touches the plane, then 
the figure formed on the plane by the wet paint (see Fig. 5) 

Fig. 5. 




will be the pattern for the cone. As the cone rolls (the 
figure represents the cone as rolling), the portion of it 
touching the plane at any instant is a slant of the cone (see 
§7.). 

(9.) Examining the figure formed by the wet paint, we find 
it to be a sector of a circle, that is, the figure contained 
between two radii of a circle and the arc they cut off. The 
length of the arc here is clearly equal to the length of the 
circumference of the base of the cone, and the radius of the 



THE TINSMITHS' PATTERN MANUAL. 



29 



aic evidently equal to the slant of the cone, v From this it is 
obvious that to draw the pattern for a cone, we require to 
know the slant of the cone (which will be the radius for the 
pattern), and the circumference of the base of the cone. 



PEOBLEM III. 

To draio the -pattern for a cone, in one piece or in several pieces^ 
the slant and diameter of the base being given. 

Pattern in one Piece. — With O A (Fig. 66) equal to the 
slant as radius, describe a lung arc ACE. What has now 



Fig 




to be done is to mark oflf a length of this arc equal to the cir- 
cumference of the base of the cone. The best and quickest 
way for this is as follows. Draw a line F B (Fig. 6a) equal 
to the given diameter of the base, and bisecl; it in G ; then 
G B is a radius of the base. From G draw G E perpendicular 



30 



T1[E TIX.'^MITIIS' PATTEKN MANlAl. 



to F B ; and with G as centre and radius G B describe from 
B an arc meeting GE in E. The arc B E is a quadrant 
(quarter) of the circumference of the "base of the cone. 
Divide this quadrant into a number of equal parts, not too 

Fia. Ga. 




many, Bay four, by points 1, 2, 3. From A (Fig. 66) mark 
off along arc ACE four parts, each equal to one of the 
divisions of the quadrant, as from A to B. Take this length 
A B equal to the four parts, that is, equal to the quadrant, 
and from B set it off three times along the arc towards E as 
from B to C, C to D, D to E. Join E to O; then A CEO 
will be the pattern required. 

Note. It must be noted that when tnis pattern is bent round to form the 

cone, the edges O A and O E will simply butt up against each other, 
for no allowance has been made for lap or seam. Let us call the junction 
of O A and O E the line of butting. Nor, further, has any allowance been 
made for wiring of the edge ACE. These most essential matters will 
be referred to immediately. 



Pattern in mobe than one Piece.-— If B be joined to 0, 
then the sector O A B will be the pattern for one-qimrter 
of the cone. If C be joined to O, then the sector O A G is 
the pattern for one-half of it. Similarly A D will givg 



THE TENSMrXHS' PATTERN MANUAL 



31 



three- quarters of the cone. A cone pattern can thus be made 
in one, two, tkree, or four pieces. If the cone is required to 
be made in three pieces, then instead of dividing, as above, 
a quadrant of the circumference of the base, divide one-third 
of it into parts,. say five ; set off five of the parts along ACE 
from A, and join the last division point to the centre ; the 
sector so obtained will be the pattern for Gne-tMrdof the cone. 
If required to be made in five pieces, divide a fifth of the 
circumferenea of the base into equal parts, and proceed as 
before. Similarly for any number of pieces that the pattern 
may be required in. 



PEOBLEM IV. 

To draw the jpatiern for a cone, the height and the diameter of 

the base heing given. 

First find the slant. OB (Fig% 7a) by P.eoblem IL Then 
with A as centre and radius A B, describe B a quadrant of 



Fig. 7a.. 



FiG. 76; 




^ Tiaduis afHasoQ 



O'O^ 



the circumference of the base, and proceed, as in Problem III. 
to draw the pattern Fig. 76 (the plain lines). » 



32 THE TINSMITHS' PATTERN MANUAL. 

ALLOWANCE FOR LAP, SEAM, WIETNG, &o. 

(10.) It has already been stated that the geometric pattera 
Fig. 6& has no allowance for seam, wiring, or edging. (Foi 
the present it is assumed that these terms are understood i, 
we shall come back to them later .on.) In the pattern 
Fig. 76 the dotted line 0' B' parallel to the edge B shows 
' lap ' for soldered searu. For a ' grooved ' seam not only must 
there be this allowance, but there must be a similar allowance 
along the edge O D. These allowances, it must be distinctly 
remembered, are always extras to the geometric pattern; that 
is to say, the junction line of D and B, or line of butting 
(see Note, Problem III ) is not interfered with. And here a 
word of warning is necessary. Suppose instead of marking off 
a parallel slip or lai3 for soldered seaHi, a slip D D' going off 
to nothing at the centre 0, is marked off, and that then, for 
soldering up, there is actually used not this triangular slip, 
but a parallel one as D D' O 0^, the result brought about will 
be that the work will solder up untrue; there will be, in 
fact, a * rise ' at the base of the work. We can understand 
the result in this way. If the parallel slip D D' 0^ used 
for soldering were cut off, there would remain a pattern 
which is not the geometric pattern, but a nondescript 
approximation, having a line of butting other than the true 
line. And it being thus to an untrue pattern that the 
parallel slip for seam is added, the article made up from the 
untrue pattern must of coUrse itself necessarily be untrue. 
In the fig. the dotted line parallel to the curve of tho 
pattern shows an allowance for wiring. For a grooved seam 
there must be on the edge O D an addition O D D' O^ similar 
to the addition on the edge O B, as above stated. 

(11.) In working from shop patterns for funnels, oil-bottle 
tops, and similar articles, workmen often find that if they 
take a good lap at the bottom, and almost nothing at the top 
of the seam, the pattern is ime. And so it is, for these 
patterns have the triangular slip D O D' added. Whereas, 
if a paralld piece D O^ O D' is uzod fox lap, tbo pattoni ig 



THE TINSMITHS PATTERN MAXIMAL. 



33 



untrue. Which again is the cas3, because, now, in addition 
to D D', an extra triangular piece D O^ is used, and this 
extra is taken off the geometric paitern. Consequently, the line 
of butting is interfered with; that is to say, the two lines 
OB and D, instead of meeting, overlap; OB forming a 
junction, with O^ IT instead of with D ; with which U B 
must always form a junction, for the pattern to he true. In 
setting out patterns, to. prevent error, the best rule to follow 
and adopt is, to first mark them out independent of any 
allowance for seams, or wiring, or edging, and to afterwards 
add on whatever allowances are intended or requisite. In 
future diagrams, allowances, where sbown, will be mostly 
shown by dotted lines. 

Definitiox. 
(12.) ri:usTUM. — If a right cone is cut by a plane parallel to 



Fig. 8a. 



Fifi. 86. 



,'- 




£ M J} 

RauUus cf Base of 
un.:zci Cona 



its base, the part containing the apes is a. complete cone, as 
QCGDL (Fig. 8a), and the part CABD containing tho 



34 



THE TINSMITHS' P ATT I UN MA'NUAL. 



base A H B K is a frustum of the cone. In otliei' words a 
frustum of a right cone is a solid having cii-cnlar ends, and 
of equal taper or inclination of slant eveiywhere hetween 
the ends. Conversely a round equally tapering body having 
top and base parallel is a frustum of a right cnne. 

(13.) Comparing such a solid with round ai tides of equal 
ta])er or inclination of slant, as pails, coffee-pots, gravy 



I 



Figs. 9. 





straiuirs, and so on (Fig. 9), it will 
be seen that they are portions 
(frusta) of right cones. 

(14.) In speaking bere of metal- 
plate articles as portions of cones, 
it must be remembered that all 
our patterns are of surfaces, seeing 
that we are dealing with metals 
in sheet ; and that those patterns when formed up are not 
solids, but merely simulate solids. It is, however, a con- 
venience, and leads to no confusion to entirely disregard 
the distinction ; the method of expression referred to is 
therefore adopted throughout these pages. 

( 15.) By Fig. 86 is shown the relations of the cone O A B 
of Fig. 8a with its portions O C D (complete cone cut oflp), and 
A B D (frustum). The portion O D is a complete cone, 
US it is the solid that would be formed by the revolution of 
the right-angled triangle OFD (both figs.) around OF. 
The triangles OFG and OFO (Fig. 8a) represent the 



THE TINSMITHS' PATTEHN MANIA K. 



35 



trian, le F D in progress of revolution. The triangle E B 
(both figs.) is the triangle of revolution of the uncut cone 
O AB (Fig. 8a) and O E H, O E A represent E B in pro- 
gress of revolution. The height of the cone OAB being 
OE (both figs.), the height of the cone O C D is OF (both 
figs.). The radius for the construction of pattern of the 
uncut cone OAB will be O B (both figs.) , for the pattern ef 
CD, the cone cut off, the radius will be O D (both figs.). 
In F E, or D M, we have the height of the frustum. Just as 
(§ 8) the portion of the rolling cone touching the plane at any 
instant is a slant of the cone, so the slant of a frustum is that 
portion of it, which, if it were set rolling on a plane, would 
at any instant touch the plane. D B is a slant of the 
frustum C A B D, The extremities of a slant of a frustum 
are * corresponding points.' Other details of coBa and 
frustum are shown in Fig. 8&. 

Fig. 10. 




(16.) It is obvious that, if the patterns for the cones A B, 
O G D (Fig. 8a) be drawn (Fig. 10) from a common centre 
O, the figure A G D B will be the pattern for the frustum 

D 2 



36 



THE TINSMITHS' PATTERN MANUAL. 



A C D B (Fig. 8a). From wbich we see that in order td 
draw the pattern for the frnstnm of a cone, we mnst know 
the slant of the con(9 of which the frustum is a portion, that 
is, we must know the radius for the construction of the 
pattern of that cone, and also the slant (radius for pattern) 
of the cone cut off. 

PROBLEM V. ^ 

Given the dimensions of the ends of a round equal-tapering body 
(frustum of right cone), and its upright height. To find the 
slant, or the height, of the cone of which it is a portion. 

Draw any two lines A, A B (Fig. 11) at right angles to 
each other and intersecting in A^ From A on either line, 

Fig. 11. 



C,^< 




of Larcje cruJL 

Bay on B A, mark off A B equal to half the diameter of the 
larger of the given ends, and from A on the other line make 
A C eq.ual to the given upright height. Draw a line C D 



tUE TINSMITHS' PATTERN MANUAL. 



37 



parallel to A B, or, which is the same thing, at right angles 
to A O, and make G D equal to half the diameter of the 
smaller endi Join B D, and produce it, meeting A in O. 
Then O A is the height of the cone of which the tapering 
body is a portion, and O B the slant. 

PEOBLEM VL 

To draw the pattern for a frustum of a cone, the diameters of the 
ends of the frustum and its upright height being given. 

The Frustum. — Draw any two lines A, B A (Fig. 12a) 
perpendicular to each other and meeting in A ; on one of the 

FiQ. 12a. 




jl Mall Longer 
Diajn£ter 



perpendiculars, eay B A, make A B equal to half the longer 
diameter (radius), and on the other make A C equal to the 
given upright height. Dfkw a line C D perpendicular to 
A O and make C D equal to half the shorter diameter. Join 
B D, and produce it, meeting A produced in 0. With A 



THE TINSMITHS' PATTERN MANUAL. 



as centre, and radius A B, describe quadrant B E, . which 
divide into any convenient number of equal parts, here 
fonr. 
To draw tho pattern (Fig. 12o) tako any point 0' as 



.-^ 




THE TINSM1T11S5' PATTERN MANUAL, 39' 

centre, and with radius OB (Fig. 12a) describe an arc 
ACE; also with same centre and radius, OD (Fig. 12a), 
describe an arc A' (J' E'. 'From any point in the outside 
curve, as A, draw a line, tlirough O', and cutting the 
inner arc in A'. From A mark off successively jjarts equal 
to those into which the qiiadrant B E (Fig. 12a) is divided, 
and the same number of ^them, four, to B. And- from B, 
along the outer curve, set off B C, CD, D E, each equal to 
A B. Join EO', cutting the inner curve in E'. Then 
A A' E' E is the pattern required. 

Just as O B (Fig. 12a) is the slant of the cone that would 
be generated by the revolution of right-angled triangle 
OAB around OA, so DB is the slant of the frustum of 
which A A' E'E (Fig. 12&) is the pattern. In the pattern 
the slant D B appears as, A A', B B', G C, &c. 

Parts' of the Frustum — If B be joined to 0', the figure A A' 
B' B will be one-quarter of the pattern of the frustum ; and 
if C be joined to 0', the figure A A' C 'C will be pattern for 
one-half of it, and so on. The paragraph *' Pattern in 
more than one Piece " in Problem III. should be re-read in 
connection with' this " Parts of a Frustum." 

(17.) The problem next following is important, in thdt, 
in actual practice, the slant of a round equal-tapering botly 
is very often given instead of its height, especially' in cases 
where the taper or inclination of the" slant is great ; as for 
instance in ceiling-shades. The only difference in the work- 
ing out of the problem from that of Problem VI. is that the 
radii required for the pattern of the bdJy are found from 
other data. Let us take the problem. 



PROBLEM VIL 

To draw the pattern for a round, equal-tapering body (frustum o/ 
right cone), (lie diameter of the ends and the slant being given. 

To find the required radii, draw any two lines O A, B A 
(Fig. 13) perpondictilar to one another, and meeting- in A 



40 



1:iiE TINSMMHS' PATTERN MANUAL. 



On either line, as A B, make A B equal to half the longer of 
the given diameters and A C equal to hajf the shorter. From 
C draw GD perjrundicular to A B. With B as centre and^ 




Half Shorter 



A^Biaineler C \B 

I ; 

W Ealf Longer Diameter > 



the given slant as radius, describe an arc cutting C I) in E. 
Join B E and produce it to meet' A O in O. Then B and 
O E are the required radii. By E F being drawn parallel to 
•A B, comparison may be made between this Fig. and Fig. 
12a, and the difference between Problems VI. and VJI. 
clearly apprehended. -To draw the pattern, proceed as in 
Pbobi,km VI. 

(18.) For large work and for round equal-tapering bodies 
which approximate to round bodies without any taper at all, 
the method of Problem VI. i>< often not available, for want 
of space to use the long radii that are necessary for the 
curves of the patterns. The next problem shows how to 
deal with such cases ; by it a working-centre and long radri 
can be dispensed, with. The method gives_fairly. good 
results. 



THE TlNSMItHS' PATTERN MANUAL. 



41 



PilOBLEM VIIL 

To draio, Without long radii, the pattern for a round equal- 
tapering body [frustum of rigid cone), the diameters of the 
ends and the upright height being given. 

First draw one-quarter of the plan. (To do this, we fore- 
stall for convenience what is taught in the following chapter.) 




Draw any two lines B 0, C (Fig. 14) perpendicular to each 
other and meeting in O. With O as centre and radius equal 
to half the longer diameter, describe an arc meeting the lines 
B 0, C in B and C. ^With as centre and radius equal to 



42 



THE TIN>!MITUS' PATTERN MANUAL, 



"Jialf tlie shorter diameter describe an arc B' C. This 
completes the one-quarte*r plan. 

Now divide B C, the largest arc, into any nnmLer of equal 
parts, say four ; and join the points of division to hy lines 
cutting B'C in 1', 2', 3'. Join 3'G, and ihroxagh 3' draw 
3'E perpendiciilar to 3' C, and equal to the given npright 
height. Join C E ; then C E may bo taken as the true ieiigth 
of 3'. Through C' draw C D perpendicular to C and 
equal to the upright height. Join C D ; then C D is the 
true length of C C. If it is inconvenient to find these iruo 
lengths on the plan, it may be done apart from it, as by 
the triangle~B P and Q. • " 

To set out the pattern. Draw (Fig. 15) any line C C 
equal to C D(rig. 14). With C and C as centres and radii 
respectively CE and 3 (Fig. 14) describe arcs intersecting 
in 3 (Fig. 15). With and C' as centres and radii respec- 
tively CE and C'3' (Fig. 14) dcsc-ribe arcs intersecting in 
3' (Fig. 15). Then and 3 ato two points in the outer 




curve of tlio pattern, and C 3' two points in the inner cuive. 
To find points 2 and 2', proceed as just explained, and with 
the same radii^ but 3' and 3 as centres instead of C and 0. 
Similarly, to find points 1' and 1, and B' and B. A curved 
line drawn from O through 3, 2, and 1 to B will be the outer 
jQurve of_oiie.- quarter of the rc(j[uired pa^ttern, and a curved 



THE TINSMITHS' PATTERN MANUAL. 



43 



line from C'througli 3', 2', and 1' to B' its inner curve; that 
is C C B' B is one-quarter of tlio pattern. Four times the 
quarter is of course the required pattern complete. 

Note. — In cases where this method wiU be most useful, the pattern is 
generally required so tl at the article can be made in two, three, four, or 
more pieces. If the pattern is required in three pieces, one-third of tlie 
plan must be drawn (see end of Problem III., p. 31) instead of a quarter, as 
in Fig. 14; tlie remainder of the construction will then be as described 
above. 

(19.) It is often desirable in the case of large work to 
know what the slant or height, whichever is not given, of 
a round equal-tapering body (frustum of right cone) will be, 
before starting or making the article. Here the following- 
problems will be of service. 



PROBLEM IX. 

•To find the slant of a round equal-tapering body (^frustum of 
right cone), the diameters of the ends and the height being 
given. 

Mark off (Fig. 16) from a point m any line O B the 
lengths of half the shorter and longer diameters, as O C, O B. 




From C draw C D perpendicular to B. Make C D equal to 
the given height, and join B D. Then B D is the slant 
required. 



44 



Tlir. 'ilNSAIlTll.S' PAT'IKRN MANUAL. 



PKOBLEM X. 

To find tJie height of a round equal-tapering body (frustum of 
right cone), the diameters of the ends and the slant being 
given. 

Mark off (Fig. 17) from a point in any line B tlie 
lengths of half the shorter and longer diameters, as in Puoblem 
IX., and from C draw C D perpendicular to O B. With B as 
centre and radius equal to the given slant, describe an arc 
cutting C D in E. Then C E is the height required. 

Essentially this problem has already been given, in the 
working of Problem VII. 



^IG 


17. 




D 


^ 


^ 



Fig 


la 




B 




\ 



B 



JL 



PEOBLEM XI. 

Given the slant and the inclination of the slant of a round equal- 
tapering body ; to find its height. 

Let A B (Fig. 18) be the slant, and the angle that A B 
makes with C A the inclination of the slant. From B let 
fall B D perpendicular to A 0. Then B D is the height 
required. 

(20.) In the workshop, the inclination of the slant of a 
tapering body is sometimes spoken of as the body being 
so many inches " out of flue." This will be explained 
in the following chapter. If the inclination of the plant 



THE TINSMFTHS' PATTERN MANUAL. 45 

is given in these terms the problem is worked thus. From 
any point D in any .lin& C A (Fig. 18) make D A equal in 
length to the number of inches the body is " out of flue," and 
draw D E perpendicular to A. With A as centre and 
radius equal to the given slant, describe an arc intersecting 
D E in B, Thou B D will be the height required. 



48 



THE TINSMITHS' PATTERN M VNl'AU 



CHAPTEK Y 

.Equal-tapeeing Bodies of which To? and Base aee 
Parallel, and their Plans. 

(21.) First let us understand what a plan is. Fi^;. 19 
jrepresents an object Z, made of tin, say, having six faces, 

Fia. Id. 




of which the A B C D and G H K L faces are parallel, as 
also the B D K n and C A L G. The ABCD and CDKL 

faces are square. The ABCD face has, soldered flat on it 
centrally, a smaller square of tin ahcd with a central 



THE TIN.^MITHS' PATTERN MANt'U.. 



47 



circular holo in it. Now suppose wires (rc'presented in the 
fig. by dotted lines), soldered perpendicnlarly to the 
A B C D face, at A, B, C, D, a, 6, c, d, E, and F (the points 
E and -'F are points at the extremities of a diameter of the 
circular hole). Also suppose wires soldered at G and H 
parallel to the other wires, and that the free ends of all the 
wii-es are cut to such length that they will, each of them, 
butt up against a flat surface (plane), of glass say, X X X X, 
parallel to the A B C D face. Lastly suppose that all tlie 
points wheie the wfres touch the glass are joined by lines 
corresponding to edges of Z (see the straight lines in tho 

Fig. 20. 




^gure on the plane) ; also that E' and F' are joined, that 
the line juiuing them is bisected, and a circle described 
pa.-sing through B' and F'. Then the complete representa- 
tion obtained is a projection of Z. Instead of actually pro- 
jecting tho points by wires, we may make tho doing of it 
another supjiosition luay, find, as if by wires, the required 
points, and diaw the projection. Tho ABCD face being, 



48 THE TINSMITHS' PATTERN MANUAL. 

eay 2 inclies siuare, the flat piece, say 1 inch squa^'e, and 
the hole ^ inch diameter, and the back face GHK L, say 
2|- in. by 2 in., then the projection that is tipon the glass 
■would be as shown in Fig. 20. The plane X XXX is here 
supposed vertical, and the projection G' C D' H' is therefore 
an elevation ; if the plane "were horizontal, the projection 
would be a plan, and we might regard A B C D as the top 
of the body, and G H K Las its base, or vice versa. We may 
define a plan then as the representation of a body obtained 
by projecting it on to a horizontal plane, by lines perpen- 
dicular to the plane. 

(22.) The plane X X X X was supposed parallel to the 
A B C D face of Z ; the plan A' B' C D' of it is therefore of 
the same shape as A B C D, and in fact A B C D may be said 
to be its own plan. Similarly the G' H' D' C is the plan of 
the back-face G H K L and is of the same ^ape as that face. 
But the plan of the face A G 11 B to which the plane is not 
parallel is by no means the same shape as that face, for the 
long edges B H and A G of the face A G H B are, in plan, 
the short lines B' H' and A' G'. We need not, however, go 
farther into this, because in the case of the bodies that now 
concern us, the hoiizontal plane on which any plan is drawn 
is always supposed to be parallel to the principal faces of 
the body, so that the plans of those faces are always of the 
same shape, as the faces. In this paragraph the plane 
X X X X is supposed to be horizontal. 

(22a.) We are now in a position to explaia the getting at 
the true length of C C in the fig. of Problem VI IL, p. 42 ; 
or, putting the matter generally, to explain the finding the 
triia lengths of lines from their apparent lengths in their 
plans and elevations. Horizontal lines being esceptedj there 
is, manifestly, for any line, however positioned in space, 'a 
vertical plane in which its elevation will appear as (if not a 
point) a vertical line. Let B E (Fig. 17, p. 44) be any line 
in the plane of the paper, and let C D be the vertical plane 
seen edgeways on which the elevation E C of B E is a vortical 
line. Then if B be a horizontal plane seen edgeways'' 



THE TINSMITUS' PATTERN MANUAL. 49 

pa.'-sing through C, the line joining the B extremity of B E 
to the C extremity of its elevation will be the plan of B E. 
We get thus the figure ECB, a figure in one plane, tlio 
plane of the paper, a right-angled triuntile in fact, of which 
the E C side is the elevation of B E, the C B side its plan, and 
the hypotenuse the line itself; a figure, which, as combining 
a line, its plan, and its elevation, we have under no other 
conditions than when t'le elevation in question is a vertical 
line. In the plane passing through E and B E, that is, in 
the plane in which these lines wholly lie, we have in the 
line that we get by joining C with B the plan, full length, 
of BE. In 1 espect of tliis plan of B E, we are cone erned 
with no other measurement, because, in a right-angled 
triangle representing a line and its plan and elevation, no 
other measurement of the plan lino can come in. Not so, 
however, with the elevation line of B E. Here other mea- 
surement of it than its length can and does come in, because 
th.it longth varies according to the position of the vertical 
plane with regard to it ; the plan length is always the same, 
But to have in the three side^ of E C B, the representation of 
B E, and its plan and elevation, it is evident that the plane 
which contains B E and its plan C B must also wholly 
contain the elevation E ; that is, the plane must be perpen- 
dicular to the piano of the triangle. Now, no matter on 
what vertical plane the line B E is projected, although the 
length of the projection will vary, the vertical distance 
between its extremities, that is, its height, never varies. 
Hence, if, in any right-angled triangle, we have in the 
hypotenuse the representation of a line, in one of its sides 
the plan of the line, and in the other side, not necessarily the 
elevation that comes out vertical, but the height of any 
elevation of the line, it comes to the same thing as if in the 
latter side we had the actual elevation that is veitical. And 
hence, further, if we have given the plan-length of an 
unknown line, and the vertical distance between its extre- 
mities, we can, by drawing a line, say C B, equal to the given 
plan-length, then drawing from one of its estremities and at 



50 THE TINSMITHS' PATTERN MANUAL. 

right angles to it, a Ime, say CE, equal to the given vertical 
distance, and finally joining the free extremities, as by BE, 
of these two lines, construct a right-angled triangle, the 
hypotenuse, BE, of which must be the true length of the 
unknown line ; for there is no other line than B E of which 
C B and C E can be, at one and the same time, plan and 
elevation. V,'e have explained this true-length matter fully, 
because we have to make use of it abundantly in problems to 
come. 

(23.) Proceeuirig to the bodies wo have- to con.-ider, wo 

Fig. 21. 




take first a frustum of a cone, Fig. 21a. To draw its plan, lot 
us suppose the estremities of a diameter of its smaller face 
top (namely points A and F of the skeleton drawing Fig. 216) 
(neither drav/ing is to dimensions), to be projected, in the 
way just explained, on to a plane parallel to the face, then, 
also as there explained, we can dra%v the circle which is a 
projection of that face. Suppose the smaller circle of Fig. 22 
to be that circle, and to be to dimensions. Projecting now, 
similarly, the extremiiies of a diameter of the larger face 
(base), namely the points G and J) of the skeleton drawing, 
on to the same plane, we can get the projection of the larger 
face. Ltjt the larger circle of Fig. 22 be that projection. 
The two circular projections will be concentriG (having tke 



The TINSMITHS* PATTHUN MANUAL. 51 

ep.me centre) because the body Z is of equal taper, and tbey 
will, together, be the plan of Z, that u Fig. 22 is that plan. 

A C and F D each show the slant, and B A and E F thn 
height. B C and D E each show the distances between the 
plans of corresponding points. 

Fig. 22. 




(24.) Turn to the skeleton drawing of Z. Here A C shovra a 
slant of the frustum (§ 15), A B its height (see J) M, Fig. 8&), 
and A and are 'corresponding points' (§ 15). Looking 
at C D E B as at the plan of the firustum, we have, in the 
point B, the plan of the point A. Joining B C, we get a 
right-angled triangle ABC; the slant A C is its hypotenuse, 
the height A B k one of the sides containing the right angle, 
and the other side ccataining the right angle, B C, is the 
distance between the plans of the corresponding points A and 
C, as also between plans of corresponding points of Z any- 
where. I'his distance is that of how much the body ie * out 
of flue ' (a workshop expression that was referred to at tho 
end of the previous chapter), in other words, how much A C 
is out of parallel with A B. What points, in the plan of 
a frustum, are the plans of corresponding points is showu 



52 



THE TIMSMITHS' PATTERN MANUAL. 



by the fig., as the line joining the plans of correspondlug 
l)oints (the line joining B and C or that joining D and E, 
for instance) will always, if produced, pass through the 
centre of the circles that constitute the plan of the frustum ; 
the centre of the circles being the plan of the apex of the 
cone of which the frustum is a part. Which leads us to 
this ; that the distance, actually, between the plans of 
corresponding points in the plan of a frustum is equal to 
half the difference of the diameters of its two circles ; for, 
the difference between E B and D is the sum of D E and 
BC, and DE and BC are equal j in other words, either 
D B or B C is half the difference betv/een E B and D G. 



Pig. 2Rrt. 




(25.J Let us now consider another equal-tajiering body 
v/hich has top and base parallel, and we will suppose it to 
have Bat parallel sides, flat ends, and round (quadrarit) 
corners. Such a body is ropresented, except as to dimen- 
Bions, in Z, Figs. 23a and 6; Fig. 236 being a skeleton 
drawing of the body represented in Fig. 23a. Extending 
our definition of ' slant ' to apply to such a body, a * slant ' 
becomes the shortest line thut can be drawn anywhere on 
the slanting surface ; and ' corresponding points ' become, 
in accordance, the extreme points of such line. Either of 
the lines F A, G B, E C, H L, or MO represent a slant of 
the body, and F and A are corresponding points ; as also 
are G and B, E and 0, H and L, and M and 0, The height 



THE TINSMITHS' PATTERN MANUAL. 



53 



of the body is represented hj eitlier of the lines F A', G B', 
EC, HL', or MO'. The plane €or the plan being parallel 
to the M P Q E face (here the top) of Z, the plan of that face 
is of the same shape as the face« The round-cornered rect- 
angle A' F' a' D' B' C of Fig. 29 is the plan to dimensions. 
For the same reason the plan of the OATS face (here the 
base) is of the same shape as that face. The round-cornered 
rectangle AF GD B G of Fig. 29 is the plan t<) dimensions. 
How actually to draw these plans we shall deal with 
presently as a probL-m. The two circles constituting the 
plan of the frustum were concentric, that is, symmetrically 
disposed with respet;t to one another, because the frustum 

Fig. 2.36. 




was an equal-tapering body ; and the plans of top and base 
of the body we are now dealing with are symmetrical to each 
other for the same reason. The two plans (Fig. 29) together 
are the plan "of the body Z. 

(26.) Looking at A B G D A' B' C' D' of the skeleton draw- 
ing (Fig. 236) as at the plan of Z,- we have, just as with the 
cone frustum, in the point A' the plan of F, in the point B' 
the plan of G, in the point C' the plan of E, in the point L' 
the plan of H, and in the point 0' the plan of M. Further as 
in the ease of the frustum, if we join any point in the plan 
of the base, as A, to the plan of its corresponding point A', 
then we have a right-angled triangle, F A A', of which the 



5-f THE TINSMITHS' PATTERN MANUA!,. 

hypotennse F A represents the slant of the 'bofly; F A', onA 
of tliG sides containing the right angle, its height, and A' A, 
the other side containing the right angle, the distance 
between the plans of the corresponding points F and A, which 
is also the distance between B and B', C and C', L and L', O 
and O', and between plans of corresponding points of the body 
anywhere, the body being of equal taper. As with Fig. 216 
what points, in the plan, are the pLxris of corresponding points 
is clear froia the fig. Wh re the plan of the body consists 
of straight lines, the plans of corresponding points are always 
the extremities of lines joining these straight lines perpen- 
dicularly; the extremities of AA', B B', C C, and L L', for 
instance. "Where the plan of the body consists of arcs, the 
plans of corresponding points (compare with cone frustum) 
are the extremities of lines joining the arcs, and which, pro- 
duced, will pass through the centre from which the arcs are 
described ; the line 0' for instance. To mate all this quite 
plain, leference shotsld again be made to Fig. 29 ; also to 
Fig. 28, which is the plan of an eq"al-taperingobody with 
top and base parallel, end having flat sides, and semicircular 
ends. In Fig. 29, A A', B B', C C, D D', are lines. joining 
the plan lines of the flat sides and ends perpendicularly, and 
the extremities of each of these lines are plans of correspond- 
ing points, that is to say, A and A' are plans of corresponding 
points, as are also B and B', C and C, and D and D'. Also 
F and F' a'^e plans of corresponding points, being the 
extremities of the line F F' which is a line joining ihe ends 
of the arcs which are the plans of one 'of the quadrant 
corners of the body. Siniilarly G and Q' are plans of corre- 
sponding points. In I'ig. 28, F F', G G', D D', E E', are lines 
joining perpendicularly the plan lines of the flat sides of the 
body at their extremities where the semicircular ends begin ; 
and F and F', G and G', D and D', E and E' are plans of 
corresponding points. ALso A and A' are plans of corre- 
sponding points, and B and B', seeing that the lines joining 
these points, produced, pass respectively through and 0', 
the centres from which the semicircular ends are described. 



THE TINSMITHS' PATTERN MANUAL? 55 

In the cone frustum, the actual distance between the plans of 
corresponding points was, we saw, ecjual to half the difference 
of the diameters of the' two circles constituting its plan. 
Similarly with the body Z of Fig. 23a, ani indeed with any 
equal- tapering body of whith the top and base are parallel, 
if we have the lengths of the top and baso given, or their 
widths, the distance between the plans of oorresponding 
points (number of inches ' out of flue ') is always equal to 
half the difference between. the given lengths or widths. 
Thus, tiie distance between the plans of corresponding points 
of Z is equal to half the difference between A B and A' B' 
(Fig. 29) or between CD and CD'; and the distance-be- 
tween the plans of corresponding points of the body of which 
Fig. 28 is tlie plan, is equal to half the difference between 
A B and A' B' of that fig., or between F D and F' D'. 

Summarising we have 

a. In the plans of equal-tapering bodies which have their 
tops and bases parallel, there' is, all round, an equal distance 
bstween the plans of corresponding points of the tops and 
bases. 

6. Conversely. — If, in the plan of a tapering boly with top 
and base parallel, there is an equal distance all lound between 
the plans of corresponding points of the top and base, then 
the tapering body is an eqiial-taporing body, that is, has an 
eqnal inclination of slant all round. 

c. The plan of a round equal-tapering body having top and 
base parallel, consists of two concentric circles. The plan of 
a portion of a round equal-tapering body having top and base 
parallel, consists of tv/o ares having the same centre. 

T'ao corncra of tha body Z (Fig. 23a) are portions (quarters) of a round 
equal-iapering body ; tlitir jilans are arcs (quadranta) of circles having tho 
sanie cuntre. 

d. Conversely. — If the plan of a tapering body having top 
and base parallel, consists of two concentric circles, then the 
body is a frustum of a right cone. Also if the plan of a 
lapsring body having top and base parallel, consists of two 



56 



THE TINSMITHS' PATTERN MANUAL. 



Fig. 21. 



arcs halving the same centre, then tlie "body is a portion of a 
frustum of a right cone. 

The plan of each enti of tho tapering body 
represented in plan in Fig. 28 consists of two ai-cs 
(semicircles) having the same centre; the ends 
are portions (halves) of a frustum of a right 
cone. The plan of each corner of the tapering 
body Z (Fig. 23a) consists of two arcs (quadrants. 
Fig. 29) having the same centre ; the coiners are 
portions (quarter.^) of a frustunr of a right cone. 
The fig. annexed represents a quadrant corner 
in plan separately. 

"We conclude the chapter with some problems. 




PROBLEM XII. 

Given the height and slant of an equal-tapering body iidth top and 
base parallel; to find the distance between the plans of 
cdrrei'ponding points of the top and base {number of inches ' out 
of fine'). 

Let A' (Fig. 25) te the given height. DraTv A'B per- 
pendicular to A' C ; with C as centre and the given slant as 




radius, describe an arc cutting B A' in A. Then A A' is the 
distance required. 



^tHE TIN.SMITH^' PATTTiRN MANUAL. 



57 



PEOBLEM XIII. 

Given the heigM of an equal-tapering body wifJi top and base 
parallel, and the inclination of slant (number of inches ' out of 
flue ') ; to find the distance hetiveen the plans of corresponding 

poirds of the top and base. 

Let G A' (Fig. 26) bo the given L&ight. Thrcragli A' draw 
a lino A' B perpeiidionlar to A' ; from any point, D, in A' B 
draw a lime D E makiug Virith A' B -an angle equal to that of 
the given inclination. From C draw CJ A parallel to E I> 
and cutting A' B in A ; then A A' is the distance required. 

Fia. 2G. 




PEOBLEM XIV. 

To drcLvo the plan of a round equal-tapenng body with top and base 

larallel [frustum of rigid cone), the diameter of either end 
oeing given and the height and slard. 

Case I. — Given the height and slant and the diameter of the 
smaller end. 

On any line O B (Fig. 17) set o& O C equal to half the 
given diameter, and from C draw D perpendicnlar to B. 



58 



THfc] TlNSMiTH.S' PATTERN MANUAL. 



Mark off C E equal to th.Q givon heigLt, arid with E as csnir^ 
and radius equal to the given slant, describe &n arc inter- 
seoting OB in B; then CBwill he the distance in plan 
between corresponding points anywhere in the frustum ; that 
is to say (by c, p, 55) C will be the radius for the plan of 
the smaller end of the frustum, and O B the radius for the 
plan of the larger end. 

Case IL — Given the height and slant and the diameter of 
the larger end. 

On any line OB (Fig. 27), set off O B equal to half the 
given diameter, and now work fn>m B towards instead ol 
from towards B ; thus. From B draw B perpendicialar 
to O B. Mark off B D equal to the given height, and with 
D as centre and radius equal to the given giant, describe an 
arc intersecting O B in E ; then B E will be the distance in 



Pig. 27. 




plan of corresponding points anywhere in the frustum ; that 
is to say (by c, p. 55) O B will be the radius for the plan of 
the larger end of the frustum, and O E the radius for the 
plan of the smaller end. 



THE TINSMITHS' PATTERN MANUAL. 59 



PROBLEM XV. 

To draio the plin of a round cqual-tapering hodi) with top andhase 
parallel {frustum of right cone), the diameter of either e id 
being given, and the 'number of inches ' out of flue ' (distance 
between plans of corresponding points). 

Case I. — Given the number of inches * out of 3ue,' and the 
diameter of the smaller end. 

The radius for the smaller circle of the plan wi^l ho half 
the given diameter; the radius for the larger circle of the 
plan will be this half diameter with the addition of the 
number of inches ' out of flue.' 

Case II. — -Given the number of inches ' out o*' flue/ and the 
diameter of the larger end. 

The radius for the larger circle of the plan will be half 
the given diameter; the radius for the smaller circle of the 
plan will be the half diameter less the number of inches 'out 
of flue.' 

(27.) It should be noted that v/ith the dimensions given 
in this problem, we can draw plan only, we could not draw 
a pattern. To do that we must also have height given, for 
a plan of small height and considerable incliiiation of slant 
is also the plan of an infinite number of other frustfi 
(plural of frustum) of all sorts of heights and inclinationB of 
sLint. 

PROBLEM XYL 

To ch'dvj the plan of an oblong eqnal-tavefing body with top and 
base parallel, C7id having flat (plans) sides and semicircular 
cT.ch. 

Gags I. — Y/here the length and width of the top are given, 
and the length of the bottom. 
Commencing with the plan of the top, we know from § 25 
that it will be of the same shape as the top ; we have there- 



60 



THE TINSMITHS' PATTERN MANUAL. 



f'jre to draw tliat sliape. On any line A B (Fig. 28) mark off 
A B equal toifco given length of the top, Frora A set cff 
AO5 and from B set off B'O' eacli equal to half the given 
width cf the top. Through O and 0' dra^ lines perpsa- 
dicular to A B ; and -with and O' as centres and O A or 
O B as radius describa a^cs meeting the perpendiculars in 
BF and EG. As DF and EG pass through the centres 
O aad 0' respectively they are diameters, and the arcs are 
Eornioircles I these diameters, moreover, are each equal to 
the given width. Join D E, F G, and the plan of the top 
is complete. 





D 


Fig. 28. 




E 






/ 


' 






/ 








\ 


A' 




ty 




61 
\ 


a" 1 


V 




1 





F 



Tlje plan of the base will be of the same shape as the 
base, and we will suppose it smaller than the top. What we 
have then to do is to draw a figure cf the same shape as the 
base, and to so place it in position with the plan of the top 
that we shall have a complete plan of the body we are 
dealing with. By a, p. 55, v/e know that the distances between 
the plans of corresponding po'cts of the top and base all 
round the f«il plan will bs equal. We have therefore first 
to ascertain the distance between the plan*? of any two 
corresponding points. This by § 26 will in the present 
instance be equal to half the difference between the given 



THE TINSMITHS' PATTERN MANUAL. Ql 

lexig'c\i3 cf the top and "base. Set off this half-diSereaca, as 
tha tase is smaller tlian the top, from A to A'. Then with 
and 0' as centresj and A' as radiua, describe the semi- 
circles D' A' F', E'B'G', Joia D' E', r G', axid w© have the' 
required plan of the hody. 

Case II. — Where the length and width of the top are given, 
and the height, and slant, or the height and the 
inclination of the slant (number of inches 'otit of 
flue'). 

First draw the plan of the top as in Cafce L Then if the 
height and .^lant are given, find by Problem XII. the distance 
batween the plans of corresponding . points. If the height 
and inclination of slant are given, find the distance" hj 
Problem XIII. If the inclination of tlio slant is givsn in the 
form of ' out of flue,' the number of inches ' out of fine ' is 
the required distance. Set off this distance from A to A' 
in the fig. of Case I., and complete the plan as ia Case I. 

Cass Hi.-— Where the length and width of the base 
(bottom) are given, and the height and slant, or the 
height and the inclination of the slant. 

On any line A B (Fig, 28) mark off A' B' equal to the 
given Isigth of the bottom. From A' set off A' and from 
B' set off B' 0' each equal to half the given width of the 
bottom. Through O and 0' draw indefinite lines D F, E G 
perpendicular to A B ; and v/ith and 0' as centres, and O A' 
as radius describe the semicircles F' A' D', G' B' E', join D' E', 
F' G*5 and we have the plan of the bottom. Xow by Problera 
XII. or Problem XIIL, as may be required, find the distance 
between the plans of corresponding points, or take the number 
of inches ' out of flue,' if this is what is given. Set off this 
distance from A' to A. With and 0' as centres and O A as 
radius describe semicircles meeting the perpendiculars 
through and O' in D and F and in E and G. Join D E, 
F Gf and the plan of the body is completed. 



62 



THE TINSMITHS' PATTERN MANUAL. 



PKOBLEM XVII. 

To draw tJie plan of an oblong equal-tapering body toiih top and 
base parallel, and having fiat sidett, fiat ends, and round 
(quadrant) corners. 

Case I.— Wtere tlie length and widtli of top and bottom 
(baso_) aro given. 

Draw any two lines A B, C D CFig, 29) perpendicular to 
each other and intersecting in 0. Make O A and B < ach 
eu ual to half the length of the top, which we will suppose 




larger than the 1 ottom, and A' and B' each equal to half 
the length of the bottom. Also make and D each 
equal to half the width of the top, and C and D' each 
equal to half the width of the bottom. Throxigh C, D, C, and D' 
draw lines parallel to A B, and through A, B, A', and B' draw 
lines parallel to C D and intei secting the lines parallel to A B. 
We have now two rectangles or oblongs, and we require 
to drav7 th round comers, which are quarters of circles. 



THE TINSMITHS' PATTERN MANUAL 63 

From tlie intersecting point E along ihe sides of the rectangle 
mark equal distances E F and E G, according to the size of 
quadrant corners required. With F and G as centres and E F 
or E G as radius, describe arcs intersecting in O' ; and with 
0' as centre and same radius, desciibe the arc F G, which 
will be a quadrant because if ihh points P and G be joined to 
O' the angle F 0' G will be a right angle (p. 21). Draw 
F F parallel to A B and G G' parallel to C B, and with O' 
PS centre and radius O' F' describe the arc F' G', which ako 
will be a quadrant. We have now the plan of one of tlio 
quadrant corners; the other corners can be drav/n in 
like manner. 

(27a.) It is important to notice that the larger corner deter- 
mines the smaller one. In practice it is therefore often bost to 
draw the smaller corner first, otherwise it may sometimes be 
found, after having drawn the larger corner, that it is not 
possible to draw the smaller curve sufiSciently large, if at all. 
To draw the smaller corner first, mark off from the intersect- 
ing point E' equal lengths E' F', E' G', according to the size 
determined on for the corner. With F' and G' as centres 
and E' F' or E' G' as radius describe arcs intersecting in O'. 
Then 0' will be the centre for the smaller corner. It will clsa 
be the centre for the larger corner, which may be described 
in simi'ar manner to the smaller corner in tho preceding 
paragraph. 

Case IL— Where the dimenoioiis of the top are given and 
tho height and slant, or the height and the inclination of 
; the slant. 

Draw the plan of the top, A D B 0. Find the diota,nca 
oetween the plans of corresponding points of the top and bace 
by Problem XII., or Problem XIII., according to what is 
given ; and set off this distance, as the base is smaller than 
the top, from A and B inwardo towards O on the line A B, 
and from D and C inv/ards tov/ards O on the line D C. 
Complete tho plan by the aid of v/hat has already bean. 
(Explained. 



64 THE TINSMITHS' PATTERN MANUAL. 

Case III. — Where tlie dimensions of tlie bottom are given, 
and tlie length and slant, or the height and the incli- 
nation of the slant. 

Braw the plan of the hottom, A' D' B' C'jfind the distance 
between the plans of corresponding points of top and bottom, 
set this off outwards from A', D', B', and C and complete the 
plan by aid of what has already been stated. 

PEOBLEM XVIII. 

To draw the plan of an oval equal-tapering body with top and 
base parallel, the length and width of the ton and bottom 
being given. 

Draw (Fig. 30) any two lines A B, C D intersecting each 
other at right angles in 0. Make A and B each equal 



Fig. 30. 




3' ■- 3 



to half the given length of the larger oval (top or bottom, as> 
may be), and O C and D each equal to half its given width. 
A B and C D v/ill be the ao:es of the <jval. From A, on A B, 



THE TINSMITHS' PATTERN MANUAL. 65 

mark off A E equal to C D the width of the oval, and divide 
E B into three equal parts. With O as centre and radius 
equal to two of these parts, as from E to 2, describe arcs 
cutting AB in Q and Q'. With Q and Q' as centres and 
Q Q' as radius describe arcs intersecting in P and P' ; and from 
P and P' draw lines of indefinite length 'through Q and Q'. 
With P and P' as centres and radius P D describe arcs (tli6 
side arcs), their extremities terminating in the lines drawn 
through Q and Q'; and with Q and Q' as centres, and radius 
Q A, describe arcs (the end arcs) to meet the extremities of 
the side arcs. This completes the plan of the larger oval. 

To draw the plan of the smaller oval. Make A' and B' 
each equal to half the length of the smaller oval, and with 
Q and Q' as centres and Q A' as radius describe the end 
curves, their extremities terminating, as do the outer end- 
curves, in the lines drawn through Q and Q' ; the point R is 
an extremity of one of tho smaller curves. With P and P' 
as centres and radius P R, describe the side curves. The 
plan of the oval equal-tapering body is then complete; of 
which either the larger or smaller ovals are plan of top and 
bottom according to the purpose the article may be required 
for. 

(28.) The platis of corresponding points in the plan of an 
oval equal-tapering body will be the extremities of any line 
joining the inner and outer curves anywhere, and that, pro- 
duced, will pass through the centre from which the curves 
where joined by the line are described. 

To draw the plan of an oval equal-tapering body with top 
and base parallel, other dimensions than the above may be 
given. For instance the top or bottom-may be given, ^nd 
either the height and slant, or the height and the inclination 
of the slant (number of inches out of flue). It will be a 
useful practice for the student to work out these cases .for 
himself by the aid of the instruction that has beeu given. 



THE TINSMITHS" PATTERN MANUAL. 



Patterns for Articles of Equal Taper or Inclination 
OP Slant, and having Flat (Plane) Sdp^faces. 

(Class I. Suhdivismi 5.) 
Definition. 

(29.) Pyramid. — A pyramid is a solid having a "base of 
three or more sides and triangular faces meeting in a point 
above tliat "base, each side of the figure forming the base 
being the base of one of the triangular faces, and the point 
in which they all meet being the apes;. The shape of the 
base of a pyramid determines its name ; thus a pyramid with 
a triangular base is called a triangular pyramid; with a 
square base, a square pyramid j with a hexagon base, an 
hexagonal pyramid (Fig. 31) ; and go on. The centre of the 
base of a pyramid is the point in which perpendicular lines 
bisecting ail its sides will intersect. If the apex of a pyramid 
is perpendicularly above the centre of its base, the pyramid 
is a rigid pyramid (Fig. 31, represents a right pyramid), in 
which case the base is a regular polygon and the triangular 
faces are a|l equal and all equally inclined. In a pyramid, 
the line joining the apex to the centre of the base is called 
the axis (the line V V, Fig. 31) of the pyramid. 

(30.) An important propertj'' that a right pyramid possesses 
is that it can be inscribed in a right cone. 

(31.) A pyramid is said to be inscribed in a cone when 
both the pyramid and the cone' have a common apex, and the 
base of the pyramid is inscribed in the base of a cone ; in 
other words, when the angular points of the base of the 
pyramid are on the circumference of the base of the cone and 
the apex of cone and pyramid coincide. 



THE TINSMITHS' PATTERN MANUAL. 



et 



(32.) Fig. 31 shows a right pyramid . inscribed in a right 
cone. The apex V is common to both pyramid and cone, and 
the A, B, C, &c., of the base of the pyramid are on the 
circumference of the base of the cone. Also the axis V V is 
common to both cone and pyramid. Further, the edges 
V A, V.B, V C, &c., of the pyramid are lines on the surface 
of the cone, such lines or edges being each a slant of the 
cone, or in other words a radius of the pattern of the cone in 




v/hioh the pyramid is inscribed. It hencs follows, that if the 
patteru of the eone in which a right pyramid is inscribed be 
set out with the lines of contact of cone and pyramid, as 
V A, V B, &C.5 on it, and the extremities of these lines be 
joined, v{e shall have the pattern for the pyramid. Thus, 
the drawing a pattern for a right pyramid resolves itself into 
first determining the cone which circumscribes the pyramid, 
and next drawing the pattern of that cone with the lines of 
contact of pyramid and cone upon it. 

■ -i 



6g THE TINSMITHS' PATTERN MANI]A=t, 

ROBLEM XIX. 

To draw the poJtern for an hexagonal right pyraifiid, its height 
and base being given. 

Draw (Fig. 32a) the plan A B C D'E F of the "base of the 
pyramid, which will be of the same shape as the base {see 
Chap. V.) ; the base in fact will be its own plan. Kext draw 
any two lines A, B A (Fig. 326), perpendicrilar to each 
other and meeting in A ; make A B equal to the radius of the 
circumscribed circle (Fig. 32a), and A equal to the given 
height of the pyramid. Join B O ; then B is a slant of the 




cone in which the pyramid can be inscribed, that is to say, is 
a radius of the pattern of that cone. The line B is also 
a line of contact of the cone, in which the pyramid can be. 
inscribed, that is, is one of the edges of the pyramid. 

To draw the pattern (Fig. 32c). With any point 0' as 
centre and B (Fig, 325) as radius, describe an arc ADA, 
and in it take any point A, Join A 0', and from A mark off 
A B, B C, C D, D E, E F, and F A, corresponding to A B, B 0, 
C D, D E, E F, and F A of the hexagon of Fig. 32a, and join 
the points B, C, D, E, F and A to 0'. Join A B, B C, C D, 
D E, E F, and F A, by straight lines ; and the figure bounded 



THE TINSMITHS' PATTERN MANUAL. 



69 



bj 0' A, the straight lines from A to A, and A 0', will be 
the pattern required. The lines B 0', C 0', &c., correspond 
to the edges of the pyramid, and show the lines on w^hich to 
' bend up ' to get the faces of the pyramid, the lines 0' A and 
0' A then butting together to form one edge, 

Similavly the pattern for a right pyramid of any numher 
of face?? can be drawn, the first step always being to draw the 
plan of the base of the pyramid ; the circle passing through 
thfi angular points of which will be the plan of the base of 
t^e cone in which the pyramid can be inscribed. 




Suppc^e, instead of the dimensions from which to draw the 
plan of the base of the pyramid, the actual plan be given. 
The centre from which to strike the circumscribing circle can 
then be found bj the Definition § 29. 

Defikitiojt. 

(33.) Truncated pyramid. Frustum of pyramid.— If a 
pyramid be cut by a plane parallel to its base, tb.6 part 
containing the apes wHl be a complete pyramid, and the 
other part v/ill be a tapering body, the top and base of which 
are of the same shape but unequaL This tapering body is. 



TO 



TUE TINSMITHS' PATTERN MANUAL. 



called a truneaicS, pjramid, or a frustum of a pyramid. Tlie 
faces of a truncated pyramid which is a frustum of a right 
pyramid are all equally inclined. In Figc 33 is shown such 
a frustum standing on a horizontal plane. 

(31.) Comparison should here he made between this defini- 



FiG. 33. 




tion and that of a frustum of a cone (see § 12), which it closely 
follows ; also between Fig. 21 and Fig. 33. 

(35.) Articles of equal taper or inclination of slant and 
having flat (plane) surfaces and top and baso parallel 
(hexagonal coffee-pots; hoods; &c.), are portions of right 
pyramids (truncated pyramids), or portions of truncated 
pyramids. 

(36.) Exactly as a pyramid can be inscribed in a cone, so a 
truncated pyramid can be inscribed in a frustum of a cone, and 
the edges of the trutjcated pyramid are lines on the surfaco 
of that frastum. The Blieleton drawing, Figo 336, shows a 
right trimcated pyramid inscribed in a cone frustum. It also 
represeKts the plan o£ the cone frustum, and that of the 
pyramid frustum, with the lines of projection (see Chapter 
v.), of the smaller end of the latter on to the horizontal 
plane. This inscribing in a cone gives an easy construction 
for setting out the pattern of a truncated pyramid ; which 
construction is, to first drav,r the pattern for the pyramid of 



THE TINSMITHS' PATTERN MANUAL. 



71 



wnich the truncated pyramid is a portion ; and tlien mark 
off on tliis pattern tlie pattern for the pyramid that is cut oflF. 
Here again comparison should be made with what has been 
stated about the pattern of a frustum of a cone (see § 16), and 
the resemblance noted. 



PROBLEM XX. 

To draw the pattern for an equal-tapering body made up of flat 
surfaces {truncated right pyramid), the height, and top and 
bottom being given. 

Suppose the equal-tapering body to be hexagonal. 

To draw the required plan of the frustum. The .plans of 
the top and bottom are respectively of the same shape as 
the top and bottom (§ 25). Draw (Fig. 34a) A B C D E F the 
larger hexagon (Problem X., Chap, II.) and its diagonals 



Fig. 3ia. 



Fig. 34&. 




A D, B E, G F, intersecting in Q. On any one of the sides of 
this hexagon mark off the length of a side of the smaller 
hexagon, as A G on A F, a,nd through G draw GF' parallel 
to the diagonal AD, and cutting the diagonal F C in F'. 
With Q as centre and Q F' as radius describe a circle. The 



7^ THE TINSMITHS' PATTERN MAHtfAL* 

points in wtich tliis cuts the diagonals of the larger ]ie"xagoB 
will be the angular points of the smaller hexagon. Join each 
of these angular points, beginning at F', to the one ne?;t 
following, as FE', E'D', &c. Then T'E'D'GIB'A' is the 
plan of the smaller hexagon, and, so far as laeed^d for our 
pattern, the plan of the ' equal-tapering body, made up of 
flat surfaces ' is complete. The lines A A', B B', G C, will be 
the plans (see and compare lines D J] and B C of Fig. 'dBb) of 
the slanting edges of the frustum. 

Next draw (Fig. 34&) two lines- O A, B A, perpendicular €o 
each other, and meeting in A ; make A B equal- to the radius 
of the circle circumscribing the larger hexagon of plan, and 

Fig, 34c. 




A C equal to the given height of the body. Through C draw 
C D peipendicular to A O, and make C D equal to the radius 
of the circle which passes through Ihe angular points of the 
smaller hexagon (Fig. 34a). Join B D, and produce it to 
meet A in O. 

To draw the required pattern (Fig. 34c). Draw any lino 
O' A, and with 0' as centre and B, O D (Fig. Bib) as radii, 
describe arcs ADG, A'D'G' (Fig. 34c). Then take thej 



THE TINSMITHS' PATTERN MANUAL. 73 

fitraight line length A B (Fig. 34a), and set it off as a chord 
from A (Fig. Sic) on the arc A D G. Do the same 
successively, from point B, with the straight line lengths 
(Fig. 34a) B C, C D, D E, E F, F G, the terminating point of 
each chord as set off, being the starting point for the next, 
the chord FG (Fig. 31c) corresponding to the straight line 
length F A (Fig. 34a). Join (Fig. 34c) the points B, C, D, E, 
F and G to 0' by lines cutting the arc A' D' G' in B', C, D', 
E', F', and G'. Join A B, B C, C D, &c., and A' B', B' G', C T>\ 
&c., by straight lines ; then A D G G' D' A' is the pattern 
required. 

The frnstum of pyramid is here hexagonal, but by this 
method the pattern for any regular pyramid cut parallel tr> 
its base can be drawn. The next problem will show methods 
for larger work. 

(370 If 0' A D G (Fig. 34c), the pattern for the cone in which the 
frustum of pyramid is inscribeii, be cut out of zinc or other metal, and 
small holes be punched at the points A, B, C, &c., and A', B', C, &c. ; and 
if the cone be then made up with the lines O' A, O' B', &c., marked on it 
inside, and wires be soldered from hole to hole successively to form the top 
and bottom of the truncated pyramid, then (1) the whole pyramid of 
which the truncated pyramid is a portion, (2) the pyramid that is cut off, 
as well as (3) the truncated pyramid, will be clearly seen inscribed in the 
cone. The making such a model will amply repay any one who desires to 
be thoroughly conversant ^ith the construction of articles of the kind now 
under consideration. 



PEOBLEM XXI. 

To draw, without using long radii, the pattern for an equal- 
tapering body made up of flat surfaces, the height and top 
and bottom being given. 

Again suppose the body to be hexagonal. 

Case I. — For ordinarily large work. 

Draw (Fig 35a) the plan as by last problem. Nest join 
A B', and draw B' G perpendicular to A B' and eqiial to the 
given height of the body. Also draw B' H perpendicular to 



74 



THE TINSMITHS' PATTERN MANUAL. 



B B' ; and eqnsiX to tlie given lieigTit. Join A G and B H ; 
then -A G and B H ate the true lengths of A B' and B B', 
respectively. 

Fig. 35a. 





To draw the patterar Draw (Fig. 355) B B' equal to BH 
(Fig. 35tt), _With B as centre and radius equal to B A 



TtiK TJNft'MITHo I'ATTKUN MANUAL,, 75 

(Fig. 35a), ami with B' as centre and radius equal to G A 
(Fig. 35a)'describe arcs intersecting in C, rigtt and left of 
BB'. ^Vith B as centre and the same GA as radius, and 
with B' as centre and radius B' A', describe arcs intersecting 
in C, right and left of B' B. 'Join B C. C C, C B' (Fig. 35b) ; 
then B B' C C is the pattern of the face B B' C G in plan 
(Fig. 35a). The other faces C D C D', B C B' C, &g, (Fig. 356) 
of the frustun) are described in exactly similar manner, G A 
being the distance between diagonally opposite points of any 
face as well as of the face B B' C C The triangles A 3' G, 
BB'H, can be drawn apart from the plan, as shown at 
K and L, 

(38.) It should bs observed that; if the pattern is truly 
Irawn, the top and bottom lines of each face will be parallel, 
18 B 0, B' C, of face B C B" C (Fig. 35 h) ; and tbat this gives 
^n easy method of testing whether the pattern has been 
•accurately drawn. 

Case II. — For very large work ; where it is inconvenient to 
diaw the whole of the plan. 

Draw AB (Fig. 36a) equal in length to the end-line of 
one of the faces of the frustum at its larger end. and produce 
it both ways. With B as centre and radius B A describe a 
semicircle, which divide into as many equal parts as the 
frustum has faces (Problem IX.). Hero it Is hexagonal, and 
the points of division working from point 1, are 2, C, 4, &c. 
Through the second division point, here C, draw a line to B, 
Ihen ABO is the angle mado in plan by two faces of the 
frastiim one with another, and A B, B G are two adjacent 
ond- lines of the plan of its larger ends. Bisect the angle 
ABC (Problem VIII.) by B S ; and draw a line C 0' from O 
making the acgie COB equal to the angle C B B 
(Probleia I.). On B C set off B D equal to the end-line of 
one of the faces of the frustum at its smaller end ; and draw 
D C parallel to BE, cutting 0' in C. Through C draw 
0' B'' parallel to B and meeting B E in B' ; and draw B' A' 
parallel to B A and eq_ual to B D or 0' B'. Join A' A, and we 



7B 



THE TINSMITHS' PATTERN MANUAI 



have in A'ABCC tbe plan of two adjacent faces of the 
tapering body or truncated pyramid. Next from B' let fall 
B' G perpendicular to A B, and make G F equal to tlie given 
height of the frustum. Join F B', then F B' is the true 
length of B'G. Through\B' draw B' H perpendicular to 
B' B and equal to the given height. Join H B, then H B is 
the true length of B B' one of the edges of the frustum. 

To draw the pattern. Draw (Fig. 366) B B' equal to H B 
(Fig. 36a). "With B and B' as centres and radii respectively 
B G and F B' (Fig. 36a) describe arcs intersecting in G 
(Fig. 366). Join B G and produce it making B A equal to 



Fia. 36&. 




F A 



B A (Fig. 36a). Through B' draw B' A' parallel to B A nA 
equal to B' A' (Fig. 36a)/ and join A A' ; then B A A' B' Tsill 
be the pattern of one face of the frustum. The adjacent i'ace 
B C C B' is drawn in similar manner. The fig. tihows the 
pattern, for two faces only of the equal-tapering body, 
because in cases where thfs method would have to be 
employed, two faces are probably the utmost tliat could be 
ciit out in one piece. Sometimes each face would have to be 
cut out separately, or perhaps even one face would have to 
be in portions. Any point in B' A' (Fig. 36a) instead of E' can 
bo chosen from "which to let fall a perpendicular to BA, andj 
the true length of B' G fpund as explained. The choice of j 
position depends upon the means at hand for drawing large) 



THE TINSMITHS' PATTERN MANUAI.: 77 

arcs, the radius of the arc B' G(rig. 3G&) increasing In length 
as the point G approaches nearer to A. If the pattern bo 
truly drawn, B A will be perpendicular to B' G. 

It iQust not be forgotten that these methods are, both of 
them, quite independent of the number of the sides of the 
py.-amid. Also it should be noted that BE does not of 
necessity pass through a division point, nor of necessity is 
C C parallel to A B. These are coincidences arising from the 
frustum being here hexagonal. 



PEOBLEM XXII. 

To drmo (he pattern fur an ohlong or square equal-taper ivg lody 
with top and base parallel, and having flat sides and ends. 
( The bottom is here taken as part of the body, and the whole 
pattern is in one piece.) 

Note.— This problem will be solveil in the problem next following. 
We adopt this course because the article there treated of is so iiiiportant 
an example of the oblong equal-tapering bodies in question, that it is 
desirable to make that, the special problem, the primary one, and this, the 
general problem, secondary to it. Its solution will be found at the end of 
Case I. 



PEOBLEM XXIII. 

To draio the pattern for a baking-pan. 

^ baljing-pan has not only to be water-tight, but also to 
stand heat ; hence when made in one piece the corners are 

seamless. 

Case L— Where the length and width of the bottom, the 
width of the top, and the slant are given. 

Draw two lines XX, Y Y, intersecting at right angles in 
(Fig. 37) ', make A' and B' each equal to half the 
length of the bottom and C and OD' each equal to half 
ita width. Through C and D' draw lines parallel to Y Y ; 
alco through A' and E' draw lines parallel to X X and inter-- 



78 



THE TINSMITHS' PATTERN MANUAL, 




Fig. ?,S. 




(QUAETB« 0? FlO, 37, ENLASGEB.) 



Tllfi TINSMITHS' I'ATTKRN MAXUAI,. ^ ( 79 

eecting the lines drawn tlirougli C and D' ; we get by this 
a rectangular figure, which is the shape of the bottom. Make 
A' A, B' B, C C, and D' D each equal to the given slant ; 
through B and A draw E F and G H parallel to X X ; through 
C and D draw S T and E P parallel to Y Y ; and make B E, 
B F, A H, and A G each equal to half the given width of the 
top. Join Q P and with Q, which is one of the angular 
points of the bottom, as centre and radius QP describe an 
arc cutting P E in P. Join Q P (the working can here bo 
best followed in Fig. 38) ; bisect the angle F Q P by Q 1? 
(Problem YIII., Chap. II.) ; and draw a line Q L making 
with P Q an aaglo equal to the angle F QM. The readiest 
way of doing this is by continuing the arc P F to L, then 
setting off F L equal to F M, and joining L Q. Now on B B' 
set oif B N equal to the thickness of the wire to be used for 
vfiring, and through N draw N F' parallel to E F (Fig. 37) 
and cutting Q L and Q P in V and F' ; make Q P' equal to 
Q ¥', and Q M' equal to Q V ; and join P' M' and M' F'. 
Repeat this construction fur the other three corners and the 
pattern will bo completed. This is not done in Fig. 37, for a 
reason that will appear presently. We shall quite under- 
stand the corners if we follow the letters of the Q comer in 
Fig. 38. These are B F F' M' P' P. The dotted lines drawn 
outside the pattern (Fig. 37) parallel to E F, P E, G H, and 
S T show the allowance for fold for wiring. It is important 
that the ends of this allowance for fold shall be drawn, as, 
fur instance, P K (both figs.), perpendicular to their respec- 
tive edges. 

Now as to bending up to form the pan. When the end 
adjacent and the side adjacent to the corner Q come to be loesit 
up on B' Q and D' Q so as to bring P Q and P Q into junctioa, 
it is evident that asP Q and P Q are equal .we shall obtain 
a true comer. To bring F Q and P Q into junction it is 
likewise manifest that the pattern will have also to be bent 
on the lines F' Q, P' Q, and M' Q. This fold on each other 
of P'QM' and P'QM' is generally still further bent round 
against Q V, 



80 'i'llli' TINSMITHS' rATTERN M.ANUAl.. 

. (39.) The trutli of the pan when completed, and the ease 
with which its wiring can be carried out, depend entirely on 
the accuracy of the pattern at the corntsrs. This must never 
be forgotten. In marking out a pattern, only one corner 
need be drawn, as the like to it can be cut out separately 
and used to mark the remaining corners by. The points 
E, S, and T, the fixing of w^hich will aid in this marking out, 
can be readily found, thus: — For the R corner to come up 
true it is clear that D R will have to be equal to D P, from 
which we learn that D P is half the length of the top. 
Having then determined the point P we have simply to set 
off D R, C S, and T each equal to DP. 

If the pattern Fig, 37 were completed with the comers as 
at R, S, and T, that would be the solution of Problem XXII.-, 
and' the ends and sides being bent up, we should get an 
oblong equal -tapering body with top and bottom parallel and 
having flat sides and ends. 

Case II. — Where the length and width of the bottom, the 
length of the top, and the slant are given. 

The only difference between this case and the preceding 
is that jy P (Fig. 37), half the length of the top, is known 
instead of B F the half- width of top. To find the half- width 
of top ; with Q as centre and QP as radius describe an arc 
cutting B F in F. Then B F will be half the width of the 
top, just as (§ 39, previous case) we saw that D P was half the 
iecgth of the top. The remainder of the construction is now 
as in Case I. 

(40.) It will be evident that, in this problem, choice of 
dimensions is not altogether arbitrary. The lines Q F and 
Q P the meeting of which forms the comer, must always be of 
the ' same length. This limits the choice ; for with the 
dimensions of bottom given, and the slant, and the width of 
the top, the length of the top cannot be fised at pleasure, 
but must be such as will bring Q F and QP into junction ; 
and vice versa if the length of the top is given. It is un- 
necessary to follow the limit with other data. 



THE TINSMITHS' PATTERN MANUAL?" 



81 



Case III. — "^'liere tlie length and width of the top, and the 
slant and the height are given. 
The data 'in this case and the next are the usual data when 
a pan has to be made to order. The difference between this 
case and tlie preceding is that the size of- the bottom is not 
given, but has to be determined from the data. Excepting 
as to finding the dimensions of this, the case is the same as 
Cases I. and II. All that we have now to do is therefore to 
find the dimensions of the bottom, thus : — 



Fig. 39. 



Fxa. 40. 





Draw OC (Fig. 39) equal to half the given length of tha 
top. and through C draw C B perpendicular to C and equal 
to the given height. With B as centre and radius equal to 
the given slant describe an arc cutting C in B'. Then Q B' 
is the required half-length of the bottom. 

Next draw O E (Fig. 40) equal to half the given width of 
the top,. and through E draw E D perpendicular to E and 
equal to the given height. With D as centre and radius 
equal to the given slant describe an arc cutting O E in D'. 
Then OD' is the required half-width of the bottom. 

Case IV. — Where the length and width of the top and the 
length ?)nd inclination of the slant are given. 

This is a modification of Case III. 

Let B B' {Vii<:. 39) be the length of the given slant, and 
the angle that B B' make with B' C be the incliv.ation of the 
slant. Through B draw B C perpendicular to B' 0. Then 
half the given length of the top less B' C will be the required 
half-length of the bottom; and similarly half the given 
width of the top less B' C will be the req^uired half-width of 
the bottom. 



82 



THE TINSMITHS' PAITKRN .MANUAL. 



PKOBLEM XXIV. 

To draw (he pattern for an equal-tapering body with top and base 
parallel, and having flat sides and ends (same as Problem 
XXIL), hut loith bottom, sides, and ends in separate pieces ; 
the length and width of the bottom, the icidth of the top, and 
the slant being given. 

To draw tlie pattern for tlie end. 

Draw B B' (Fig. 41) e(|ual to the given slant, and throtigh 
B and B' draw lines C D and C D' perpendicular to B B'. 
Make B C and B D eacli equal to lialf the given width of the 
top ; also make B' C and B' D' each equal to half the given 
width of the bottom. Join C C and D D' ; thea C'GBW 
will be the end pattern required. 



Fig. 41. 



Fig. 42. 




End Patters 



Side PATTERy. 



To draw the pattern for the side. 

Draw E E' (Fig. 42) equal to the given slant, and through 
E and E' draw F G and F' G' perpendicular to E E'. Make 
E' F' and E' G' each equal to half the given length of the 
bottom, and with F' and G' as centres and radius D'D 
(Fig. 41) describe arcs cutting F G in points F and G. Join 
F'F and G'G; then F'FGG' will be the side pattern 
required. 

It should be noted that, as in the preceding problem, the 
width of the top determines the length of the top and mce 
uersa, also that lines such as D D', G G' must be equal. 



THE TINSMITirS' PATTEKN MANUAL. 8.'^ 

(41.) If the ends of tlie body are seamed (• knocked up ') 
on to the sides, as is usual, twice the allowance for lap shown 
in Fig. 41, must be added to the figure C C D' D. Similarly 
if the sides are seamed on to the ends, a like double allowance 
for lap must be added to F F' G' G, instead of to the end 
pattern. 



84 tllE TINSMITHS' PATTERN MANUAL. 



CHAPTER VII. 

Patterns for Equal-tapeking Aktioles op Flat and 
Curved Surfaces combined. 

(Class I. Subdivision c.) 

From what has been stated about the plans of equal- 
tapering bodies, and fronid, p, 55, it will be evident that the 
curved surfaces of the articles now to be dealt with, are 
portions of frusta of cones. 

PROBLEM XXV» 

To draw the pattern for an eqnal-inperiiig body with top and 
bottom parallel, and having flat sides and equal semicircular 
ends {an '■equal-end' pan. for iiisfance), the dimensions of 
the top and bottom of the body and its height being given. 

Four cases will be treated of ; three in this problem and 
one in the problem following. 

Case I. — Patterns when the body is to be made up of four 

pieces, 
f 

We may suppose the article to be a pan. Having drawn 

(Fig. 43) the plan AECSA'D'C'B' by the method of 

Problem XYI. (as well as the plan, the lines of its construction 

are shown in the fig.), let us suppose the seams are to be 

at A, B, C, and D, where indeed they are usually placed. 

Then we shall require one pattern for the flat sides and 

another for the curved ends. 

To draw the pattern for the sides. 

Anywhere in AD (Fig. 43) and perpendicular to it draw 



THE TINSMITHS' PATTERN MANUAL. 



S5 



FT, an3 make FH equal to the given height ;• join F'H; 
then F'H is the length of the slant of the article, and 
therefore the^ividth of the pattern for the sides. 




Draw A A' (Fig. 44) equal to F'H (Fig. 43). Through 
A and A' draW lines perpendicular to A A'. Make A D 
(Fig. 44) equal to A D (Fig. 43), and draw D IK parallel to 

Fig. 44. 




Side Pattekn. 



A A'. The rectangle ADD' A' will be the side pattern 
required. The fig. also shows extras for lap. 

To draw the pattern for the ends. 

Draw (Fig. 46a) two lines D A, .0 A, perpendicular to ©ne 
another and meeting in A ; make D A equal to D (Fig. 43), 
the larger of the radii of the semicircular ends ; and on A O 
set off A G- equal to the given height. Draw G D' perpen- 
dicular to A G and equal to D' (Figi 43), the smaller of the 
radii of the semicircular ends ; join D D' and produce it, 
meeting A in 0. Then with any point 0' (Fig. 456), as 



86 



-THE TINSMITfTS' PATTERN MANUAL. 



centre, and D (Fig. 45a) as radius, describe an arc D C, and 
with the same centre and radius D' (Fig. 45a) descrihe an 
arc D' 0'. Draw any line D O', cntting the arcs in points 
D and D'. Divide D E (Fig. 43) into aoy number of equal 
parts, say three. From D (Fig. 45&) mark off these three 



Fig. 45w. 



Fig. 456. 








dimensions to E, make E C equal to D E, and join C 0' ; then 
D C C D' will be the end pattern required. 

If R he joined to O', then D D' E' R will be half the pattern. 
The centre line E E' is very useful, because, in making up 
the article, the point E' must meet the lino SE (asfs) of 
Fig. 43, otherwise the body will be twisted in consequence 
of fche bottom not being true with the ends. 

Case II. — Pattern when the body is to be made up of twrt 

pieces. 

Secondly, suppose the article can be made in two pieces 
(halves), with the seams at E and G (Fig. 43) , the line EG 
(part only of it shown) being the bisecting line of the plan. 
It will be seen hj inspection of the fig. that we require one 
pattern only, namely, a pattern that takes in one entire end 
of the article with two half-aides attached. 



■IIIK TINSMITHS' PATTKHN MAJfUAL 



87 



Draw tlia end pattern DD'C'C (Fig. 46) in precisely the 
same manner that it is drawn in Fig. 45&. Throngh C and C 
draw indefinite lines G C G', perpendicular to C C, and 
through. D and D' draw indefinite -lines D E, D' E' perpen- 
dicular to D D'. Make D E, D' E' each equal to D E (Fig. 43) 

Fig. 46. 




V 

and join E E' ; also make C Gr, C G', each equal to ( 
(Fig. 43), and join G G'. Then E E' G' G will be the patt( 
required. 



CG 

tern 



Gas3 III. — Pattern when the body can be made of one 
piece. 

Thirdly, suppose the article can be made in one piece, with 
the seam at S (Fig. 43). Then evidently the pattern will be 
made up of the pattern of one entire end, side patterns 
attached to this, and half an end pattern attached to each 
side pattern. 

Praw DD'C'C (Fig. 47) the end pattern. Through 
C and C draw lines perpendicular to C C, and each equal to 
C B (Fig. 43) ; and join B B'. Produce B B', and with B as 
centre a,nd D (Fig. 45a), the larger of the radii for the end 
pattern, as radius;, describe an arc cutting the pruduced lin© 



SB 'HIE TlX.SMrinS fAffEKN M\Xt"AI.. 

in 0« With as centre and B and O B' as radii respec- 
tively, draw arcs B S, B' S'. Make B S eqvial to D -E (Fig. 47) 




■-.69 



and join SO. Repeating tliis coustrnction for D D' A' A the 
other side pattern, and AA'S'S the remaining half-end 
pattern, will complete the pattern req^uired. 



nii; TlN.^MJJliS' I'ATI.KRN MANIAL 



89 



PEOBLEM XXVI. 

To draw, without long radii, the pattern for an equal-lnxiering 
hody with top ond bottom parallel, and having jUit slides and 
equal semicircular ends ; the dimensions of the top and 
hutiohi of the body and its height being given. 

TLis proLlem is a fourth case of the preceding, and 
exceedingly useful where the work is so large that it is 
inconvenient to draw the whole of the plan, and to use long 
radii. 

Draw half the plan (Fig. 48a). Divide D C into six or 
more equal parts, and join 1, 2, &c., to 0, Ly lines cutting 
D' C in 1', 2', &c., and join D 1'. Draw- D E perpendicular to 



Pig. 48a. 



Fio. 486, 



^^--..^ 


V"~->v^ 




2» 


A" 


\ 





y 


i 




\ 



S ^/ A 




]>' > 



D 1' and equal to the given height, and join E 1'. (The line 
1 to 1' appears to, but does not, coincide with E 1'.) Then 
E 1' may be taken as the true length of D 1'. Nest, producing 
as necessary, make D A equal to the given height. Joining 
D' to A gives the true length of D D'. 

To draw the 'end pattern. Draw Fig. (486) D D' equal to 
D'A (Fig. 48a). With D and D' as centres, and radii 
resjpectiyely D 1 and E 1' (Fig. 48a) describe arcs intersectingf 



95 THE TJNSMitllS' PATTERN MANUAL. 

in point 1 (Fig. 48&). With D' and D as centres and radii 
respectively D' 1' and E 1' (Fig. 48a) describo arcs inter- 
secting in point' 1' (Fig. 486). By using points 1 and 1' 
(Fig. 486) as centres, instead of D and D', and repeating tlie 
constructiou, tlie points 2 and 2' can be found. Next, using 
points 2 and 2' as centres and repeating tlie construction, 
find points 3 and 3', and so on for the remaining points 
necessary to complete the end pattern, which is completed 
by joining the various points, as 3 to 3' by a straight line ; 
D, 1, 2, and 3, by a line of regular curve ; and D', 1', 2' and 
3', also by a line of regular curve. Only half the end 
pattern is shown in Fig. 486. The side pattern can bo 
drawn as shown in Fig. 44. 



PEOBLEM XXVII. 

To draw Hie pattern for an equal-tapering hody with top and 
bottom parallel, and having flat sides and ends, and round 
corners {an oblong pan with round corners, for instance') ; 
the height and the dimensions of the tt>p and bottom being 
given. 

Again four cases will be treated of; three in this problem, 
tod one in the problem following. 

Case I. — Pattern when the body is to be made un of four 

pieces. 

The plan of the article, with lines of construction, drawn 
by the method of Problem XVII., is shown in Fig. 49. We 
will suppose the seams are to be at P, S, Q, and K, that is, at 
the middle of the sides and ends. The pattern required is 
therefore one containing a round comer, with a half-end and 
a half-side pattern attached to it. The best course to take 
is to draw the pattern for the round comer first, which, as 
will be eeeu from the plan, is a quarter of a frustum of a 
.cone. 



TIIL; TINSMlTIIci' PATTERN MANUAL. 



91 



Draw two lines CFi^. 50a) OA, C A, meeting perpendi- 
cularly to one another in A ; make A C equal to O A (Fig. 49) 
the radius of the larger arc of the plan of one of the corners, 




fiTid make A D equal to the given height of the hody. Draw 
1)E p(!rpendii'nlar to A D and equal to the radius of the 
smaller arc of the plan of a corner, and join C E and produce 



Fiu. bO't, 



Fig. 506. 



-o 



■S B 



J" - - 




it to peet AO in 0. With any point 0' (Fig. S05) as centre 
and radius equal to O C (Fig. 50rt), describe an arc A B, and 
with the sajue cunire and raJins equal to O E (Fig. 5Qa) 
dcsoiibe an arc A'B'. Dr^jW any line AO' cutting iho arcs 



^2 



THE TlN.SMITllS' PATTERN MANUAL. 



in the points A and A', and make A B equal to A B (Fig. 49) 
"by marking off the same numher of equal parts along A B 
(Fig. 50b) that we divide (arLitrarily) AB (Fig. 49) into. 
Join B O', cutting in B' the arc A' B'. Through B and B' 
draw B P, B' P' perpendicular to B B', make B P equal to B P 
(Fig. 49) and draw P P' perpendicular to B P. Through 
A and A' draw A B and A' R' perpendicular to A A', make 
A R equal to A R (Fig. 49) and draw R R' perpendicular to 
A R. Then R R' P' P will he the pattern of one round comer, 
with a half-end and a half-side pattern attached left and 
right. 

Case II.— Pattern when the body is to bo mado up of .two] 

pieces. 

I Now suppose the seams are to be at the middle ^f each 
end, at R and S (Fig. 49). The pattern" reqiiired will then 
be of twice the amount sh'^wn in Fig. 505. It will be found 
best to commence with the side pattern. 

Draw in the plan (Fig. 49), any line X X perpendicular to 
the Q side-line ; then X and X' will be plans of corre- 
sponding points (§ 26). Make XF equal to the given 

Fig. 51. 




height of the body and join X' F. Then X' F is the length 
of a slant of the body. Draw B B' (Fig. 51) equal to X' F 
(Fig. 49), and through B and B' draw B C and B' C perpen- 
dieiilar to B B'. Make B C equal to B (Fig. 49) and draw 



THE TINSMITHS' PATTERN MANUAI-. 93 

C C perpendicular to B C ; then B C C B' will be the pattern 
for the side. 

We have now to join on to this, at B B' and C C, the 
patterns for the round corners, which can be done. thus. 
Produce B B' and C C',.and make B O' and C O" each equal to 
O C (Pig. 50a) the radius of the larger arc of the corner 
pattern. With O' and O" as centres and O'B as radius, 
describe arcs B A and C D, and with the same centres and 
O'B'' as radius, describe arcs B'A' and CD'. Make BA and 
C D equal each to B A in the plan (Fig. 49), and join A C 
cutting the arc B' A' in A', and D O" cutting the arc C D' in 
D'. Throtigh A and A' draw A R and A' E' perpendicular to 
A A' ; make A E equal to A E (Fig. 49), and draw E E' per- 
pendicular to A E. Next through D and D' draw D S and 
D' S' perpendicular to D D' ; make D S equal to D S (Fig. 49), 
and draw S S' perpendicular to D S. Then E W S' S will be 
the pattern lequired. 

It should be noted that O'B' and 0"C' should be each 
equal to E (Fig. 60a), or the pattern will not bo true. 
This gives a means of testing its accuracy. 

Case III.— Pattern when the body is to be made up of one 

piece. 

When the body is made in one piece it is usual to have the 
seam at S (Fig. 49). .It is now best to commence with the 
end pattern. 

Draw A A' (Fig. 52) equal to X'F (Fig. 49), which, the 
body being equal-tapering, is the length of its slant any- 
where, and draw A H and A' H' perpendicular to A A'. 
Make A H equal to AH (Fig. 49) and draw H H' perpendi- 
cular to A H ; then A A' H' H will be the end pattern. Next 
produce A A', and make A 0' equal to O (Fig. 50a). Then 
O' will be the centre for the arcs of the corner pattern, which, 
drawn by Case II. of this problem, can now be attached at A A'. 
Through B and B' draw B C and B' C perpendicular to B B' ; 
make B C equal to B C (Fig. 49), and draw Q C. perpendicular 
to B 0. That completes the addition of the side pattern 



94 



THE TINSMITHS' PATTERN MANUAL. 



B B' C C to the corner pattern. Nest produce C C, mark tho 
necessary centre, and attach to C C the comer pattern 
CDD'C, just as ABB' A' was attached to A A'. Then 
through D and D' draw D S and D' S' perpendicular to D D' ; 
make D S equal to D S (Fig. 49) and draw S S' perpendicular 



Fio. 52. 




to D S ; this adds half an. end-pattern to D D'. By a repeti- 
tion of the foregoing working on the H H' side of the end- 
pattern H H' A' A, the portion H H' S' S on the H H' side may 
be drawn, and the one-piece pattern S H A S S' A' H'" S' of tho 
body we are treating of completed. 



PBOBLEM XXYin, 

To dremy w?,tli©nt long radii, the pattern fcv an equal- 
tapesring hcd'g with top and hoitom parallel, and having flat 
eides and ends, and round corners ; the Imght and the dimen- 
sions of the top and bottom being given. 

This is a fourth case 6f the preceding problem, and we 

will apply it to the first. case of that problem", that is, when 



THE T1X.SM1TI1.S- l-AfTERN .MANUAL. 



95 



the body is to be made up of four pieces ; now, however, 
assuming that the pieces are so large that their patterns 
cannot be conveniently drawn by the method there given. 

Draw (Fifj. 53a) V S S' F, one-quarter of the plan of the 
body. Divide C D the larger arc of the plan of the round 
comer, into three or more equal, parts, and join each division- 
point to O the centre of the arcs of the plan. Draw D F 
perpendicular to D D' and equal to the given height of the 
body, and join D'F; then D' F will be the true length of 
DD'o Next join D2'; draw DE peroendicular to it and 

Fig. 586. 





equal to the given height, and join 2' E ; then 2' E may be 
taken as the true length of D 2', (In the fig. a portion of 
the line 2' E coincides with the line from 2 to 2'. This is, 
however, merely a coincidence cf the particular case.) 

Now draw (Fig. 536) D D' equal to D' F (Fig. 53a), and 
•with D and D' as centres and radii respectively D 2 and 2' E 
{Fig, 53a) describe arcs intersecting in point 2. Also with 
D' and D as centres and radii respectively D'2' and 2'B 
describe arcs intersecting in point 2', Kepeating this con- 
struction, but using points 2 and. 2' just found as' centres 
instead of D and D', the points 1 and 1' can be found. Simi- 
larly using points 1 and 1' as centres and repeating the 
construction we can find the points G and C. Join C C ; 
draw a regular curve from C through 2 and 1 to D, and 
another regular: curve from D' through 2' and 1' io C ; then 
in DCC^D' we have the Datlern of the corner. To this 



96 THE TIN.SMITllS- PATTERN MANUAL. 

corner pattenl, to get at the pattern we require, we have to 
attach a half-end and a half-side pattern. 

Draw D S perpendicular to D D' and equal to D S (Fig. 53a). 
Also, draw S S' perpendicular to D S, and through D' draw 
D' S' parallel to D S. The half-end pattern is now attached. 

Through C and C draw P and C P' perpendicular to 
CC. Make CP equal to CP (Fig. 53a), and draw PF 
perpendicular to P. This attaches the half-side pattern, 
and completes the pattern required. 

If the hody is to h^ made in more than four pieces it will 
still be generally possible to have a complete corner in one 
of the pieces. The corner should always be marked out 
first, and whatever has to be attached should be added as 
described. 

PKOBLEM XXIX. 

To draw the pattern for an ovah equal-tapering "body with top and 
bottom parallel, the height and the dimensions of the top and 
bottom being given. 

Again we deal wit!)., four cases ; three in this problem and 
one in the following. 

Case L— Patterns when the body is to be mad© up of four 

pieces. 

The plan of the body with lines of construction, drawn by 
Problem XVIII., is shown in Fig. 54. It will be evident from 
the plan and from d, p. 55, that the body is made up of 
two round equal-tapering bodies (frusta of cones) ; the ends 
C D D' C, E E E' ¥', being equal portions of a frustum of one 
cone, and the sides, E G C E', D F F' D', equal portions of a 
frustum of another cone. Such being the case, it is clear 
that we require two patterns, one for the two ends, and the 
other for the two sides ; also that the seems should be at 
E, C, D, and F, where the portions meet. 

To draw (Fig. 555) the ends' pattern. 

From any point A draw two linGS A, B A (Fig. 55a) 



TEE TiNSMITHS' PATTERN MANUAL. 
Fjr,. r>l. 



S A 




1)8 



THE TINSMITHS' PATTERN MANUAL. 



Fig. 56a. 







THE TINSMITHS' PATTERN MANUAL. 99 

perpendicular to each otKer ; make A B equal to Q B (Fig. 54) 
tho radius of the larger cnd-cUrve in the plan, and A D equal 
to the height of the body. From D draw D E perpendicular to- 
A O and equal to Q B' (Fig. 54) the radius of the smaller end- 
curve of the plan ; join B E and produce it, nieeting A in 0. 
Then with any point O' (Fig. 555) as centre, and B (Fig. 55a) 
as radius, describe an arc C D ; and with same centre and E 
(Fig. 55a) as radius describe an arc C D'. Join any point 
C in the arc C D to 0' by a line cutting the arc C D' in C ; 
divide the arc C D (Fig. 54) into any number of equal parts, 
and set off from C (Fig. 55i>) along C D parts equal to and as 
many as C D (Fig. 54) is divided into. Join D 0' cutting the 
arc C D' in D' ; then D D' C C will be the pattern required. 

To draw the pattern for the sides. 

From any point A draw two lines A C, A perpendicular 
to each other, and make A G equal to P C (Fig. 54), the 
radius of the larger side-curve and A B equal to the given 
height of the body. From B draw B E perpendicular to 
A B and equal to P C, (Fig. 54) the radius of the smaller 
side-curve. Join C E and produce it, meeting A in O. 
With any point 0' (Fig, 56fe) as centre and O C (Fig. 56a) 
as radius describe an arc C E ; and with same centre and 
O E (Fig. 56a) as radius describe an arc C E'. Now join 
C O', and, proceeding exactly f^imilarly as in drawing \the 
ends' pattern, complete E E' C C the pattern for the sides. 

Case II.— -Patterns when the body is to be made up of two 

pieces. 

In this case the seams are usually put at the ends A and B 
(Fig. 54). It is evident that one pattern only is now retiuired, 
and that this is made up of a side pattern, with, right and 
luft attached to it, a half-end pattern. 

Draw (Fig. 57) a side pattern ECC'E' in tho way 
described in Case I. Then with C as centre and the radius 
of the ends' pattern, that is B (Fig. 65a) as radius, set off 
the distance C 0. With as centre and radii 00 and OC 
describe arcs C B and C B'. Make C B equal to C B (Fig. 54) 



100 



TilE TINSMITHS' PAttEUN JfAKL'At. 



and join B 0, cutting C B' in B'. Then C B B' C is a lialf-enrl 
pattern attached to C C. The half-end pattern E A A' E' is 



Fra. 57. 




added at E E' by repeating the construction just described. 
This completes A B B' A' the pattern required. 

C.SJSE III. — Pattern when the body is to be made up of One 

piece. 
We will put the seam at A (Fig. 54), the middle of one 
end. Draw C D D' C (Fig. 58), an end pattern in the way 
described in Case I. Produce C C and make C equal to 



I'liE TINSMITHS' PATTERN MVXUAU 



loi 



6 (Fig. 56a), the radins for a side pattern. With as 
centre and radii C and C descriha arcs C E, C i)'; Make 







V 



C E equal to CE (Fig. 54) and join E cutting C'E' in E'. 
Make E 0' ecj[Tial to B (Fig. 55a) and witii O' B as radius 



10^ 



THE TINSMITHS' PATTERN MANUAL. 



describe an arc E A. With same centre and O' 1? as radiua 
describe an arc E' A'. Make E A equal to E A (Fig. 54) and 
join A 0', cutting E' A' in A', The remainder DAA'D' of 
the pattern can be drav/n by repeating the foregoing con- 
Btraction. The figure A C D A A' D' C A' wiU be the pattern 
required. 

PSOBLEM XXX» 

To draic, without long radii, ilie paiiern for an <ysal eqiial" 
tapering body witli top and hoitom parallel, the height and 
the dimensions of the top and bottom being given. 

This is a problem v/hich will be found very useful for 
large work, especially with the pattern for the sides, the 
radius for which is often of n)ost inconvenient length. 
3) 



Fig. 5'Ja. 




■•2' J' 

End Paiterk. 



To draw tho ends' pattern. 

Draw (Fig. 59«) the plan CDD'C of the end of the body. 
Divide G D into f«jur or more eq[ual parts, and join, 1, 2. &c. 



i'llE TINSMITHS- PATTERN MANUAL: 



103 



tc Q, by lines cutting C D' in 1', 2', &c. From C draw C D 
perpendicular to C C and equal to the given height. Join 
C D, then C D is the true length of C C. Join C 1' ; draw 
V E perpendicular to it and equal to the given height ; and 
join C E. Then C E may be taken as the true length of C 1'. 
Draw C C (Fig. 596) equal to C D (Fig. 59a}. With C and 
C as centres and radii respec'ively equal to C 1 and CE 
(Fig. 59a) describe arcs intersecting in point 1. With C and 
C as centres and radii respectively equal to C 1' and C E 
(Fig. 59a) describe arcs intersecting in point 1'. By using 
points 1 and 1' as centres and repeating the construction, 
points 2 and 2' can be found. Similarly find points 3 and 3', 
and D and D'. Join D D', draw a regular curve from 
through 1, 2, and 3, to D, and another regular curve from C 
through 1', 2', and '3', to D'. The figure G C D' D is the 
pattern required. 

f IQ, 60a. 



Fig. GOb. 




Side Pattern. 



To draw the sides' pattern. 

Draw (Fig. 60a) the plan E C C E' of the side of the body. 
Divide E C into four or more equal parts, and join 1, 2, &c., 
to P, by lines cutting E' C in 1', 2'^ &c. From E draw E F 



-104 THK TINSMITHS' PATTERN MANUAL. 

perpendicular to E E' and equal to the given height. Jcin 
E' F ; then E' F is the true length of E E'. Next join E 1' ; 
draw 1' D perpendicular to E 1' and equal to the given 
height, and join E D ; then E D may be taken as the true 
length of E 1'. Now draw (Fig. 606) E E' equal to E' F 
(Fig. 60a), aud with E and E' as centres and radii respec- 
tively equal to E 1 and D E (Fig. 60a) describe arcs inter- 
secting in point 1. With- E' and E as centres and radii 
respectively equal to E' 1' and D E (Fig. 60a) describe arcs 
intersecting in point 1'. Eepeating the construction with 
points 1 and 1' as centres, the points 2 and 2' can be found ; 
and similarly the poiuts 3 and 3' and the points C and C 
Join C C ; draw a regular curve from E through 1, 2, and 3, 
to C, and another from E' through 1', 2', and 3', to C. The 
figure E E' 0' C will be the pattern required. 



THE TINSMITHS PATTERN MANUAL. 105 



Book II. 

CHAPTER L 

Patterns for Eound Articles of Unequal Taper or 
Inx^lination of Slant. 

(Class II. Suhdivision a.) 

(42.) We stated at the commencement of the preceding 
division of our subject that it was advisable, in order that 
the rales for the setting out of patterns for articles having 
equal slant or taper should be better understood and remem- 
bered, to consider the principles on which the rules were 
based ; and the remark is equally true and of greater im- 
portance in respect of the rules for the setting out of patterns 
for articles having unequal slant or taper. We have shown 
that the basis of the rules for the setting out of patterns for 
equal tapering bodies is the right cone ; we purpose showing 
that the basis of the rules for the setting out of patterns for 
articles of unequal slant or taper is what is called the oblique 
cone. The consideration of the cone apart from its species, 
that is, apart from whether it is right or oblique, which 
becomes now necessary, immediately follows. 

Definition. 

(43.) Cone. — A cone is a solid of which one extremity, tbo 
base, is a circle, and the other extremity is a point, the apes. 
The line joining the apex and the centre of the base is the 
axis of the cone. 

(44.) Given a circle and a point in the line passing through 
the centre of the circle at right angles to its plane. If au 
indefinite straight line, passing always through the given 
point, move through the circumference of the given circle, 



io() 



5'HE TINSMITDS- PATTERN MANUAL. 



ttere will lie tbereby generated bet-ween tho point and tlie 
circle, a solid ; this solid is the right cone. 

(45.) If, all other conditions remaining the same, the given 
point is out of the line passing through the centre of the 
circle at right angles to its plane, the solid then generated 
will be the oblique cone. 

(46.) Figs. 1 and 2 represent oblique cones. The lines 



Fig. 1 




YA, Y C, VF, and VG (either fig.) drawn from the apex 
V to the circumference of the base of the cone, are portions of 
the generating line at snccessive stages of its revolution. It 
is but a step from this and will be a convenience, to regard 
these- lines, first, as each of them part of an independent 
generating line, and then as each of them a complete 
generating line. Shaded representations of various oblique - 
cones will be found later on. 

(47.) Comparing now the right and oblique cones. A 
right cone may be said to be made up of an infinite number 
of equal generating lines, and an oblique cone of an infinite 
number of unequal generating lines. 

(48.) "If a right cone is placed on a horizontal plane, the 
apex is vertically over the centre of the base. 

(49.) If when a cone is placed on a horizontal plane the 
apex is not vertically over the centre of the base, the cone is 



THE TINSMITHS' PATTERN MANUAL. 107 

olbUqtie. Hence all cones not right cones are oHique cones. 
Hence also a cone is oblique if its axii is not at right angles 
to every diameter of its base. 

(50.) In the right cone, any plane containing the axis is 
perpendicular to the base of the. cone, and contains two 
generating lines (see Figs, la and '2a, p. 25), 

(61.) la the oblique cone, only one plane containing the 
axis is perpendicular to the base of the cone, and this plane 
contains its longest and shortest generating lines. In Fig, 
1 or 2 the lines V A and Y G are respectively the longest and 
shortest generating lines, and the plane that contains these 
lines contains atSo V O the axis of the cone. 

(52.) The obliquity of a cone is measured by means of the 
angle that its axis makes with that radius of the base that 
terminates in the extremity of the shortest generating line. 
Thus the angle V G gives the obliquity of either cone V A G. 
The angle V G also gives the inclination of the axis. 
As the angle V A G is in the same plane as V G and 
smaller than that angle, it will be seen that not only is Y A 
the longest gon.erating line, but it is also the line of greatest 
inclination on. the cone. Similarly as the angle that Y G 
makes with A G produced is greater than YOG, the line 
Y G is not only the shortest generating line, but also the 
line of least inclination on the cone. 

(S3.) The plane that contains the longest and shortest 
generating lines bisects the cone; consequently the generating 
lines of either half are, pair for pair, eqtial to one another. 

(54.) If the elevation (see § 21, p. 48) of an oblique cone 
be drawn on a plane parallel to the bisecting plane, the 
elevations of the longest and shortest generating lines will 
be of the same lengths respectively as those lines. Thus if 
the triangle Y A G (either fig.) be regarded as the elevation 
of the cone represented, then Y A will be the true length of 
the longest generating line of the cone, and Y G of the 
shortest. In speaking in the pages that follow, of elevation 
with regard to the oblique cone, we shall always suppose it 
to be on a plane parallel to the bisecting plane. 

^55.) If the hypotenuse of a right-angled triangle represent 



108 TUE TINSMITHS' PATTERN MANUAL. 

a generating line of any cone, riglit or oUIque, then one of 
the sides containing the right angle is equal in length to the 
plan of that line, and the other side is equal to the height 
(distance between the extremities) of any elevation of it. 
^ee, p. 25, the triangles B A, G A, O E A, D C, O K C, 
O F G, and, p. 109, the triangles V A V, V B V, V V, &c. 
See also § 22a. 

(56,) As the generating lines of the oblique cone vary in 
length, the setting put of patterns of articles whose basis is 
the oblique cone (that is to say, the development of the 
curved surfaces of such articles) differs from that apper- 
taining to articles in which the right cone is involved. The 
principles, however, are the same in both cases. In develop- 
ments of the right cone, if a number of its generating lines 
are laid out in one plane, then as they are all equal and have 
a point (the apex) in common, the curve joining their 
extremities is an arc of a circle of radius equal to a generating 
line ; while in developments of the oblique cone, as the 
generating lines, though they still have a point in common, 
vary in length, the individual lengths of a number of them 
must be found, as well as the distances apart of their 
extremities, before, by being laid out in one plane at their 
proper distances apart, a curve can be drawn through their 
extremities, and the required pattern ascertained. 

The problem which follows gives the working of this in 
full 

PEOBLEM I. 

To draw the pattern (develop the surface) of an oblique cone, the 
inclination of the axis (§ 52), its length and ike diameter 
of the base of the cone being given. 

Eirst (Fig. 3) draw V A G the elevation of the cone, and 
KdG half the plan of its base ; thus. Draw any line X X, 
and at any point in it, make the angle VOX equal to 
the given inclination of the axis (a line V is omitted in 
the fig. to avoid confusion), and make V equal to the 
given length of the axis. With as centre and half the 



THE TINSMITHS' I'ATTERN ^MANUAL 



109 



given diameter of tlie base as radius describe a Bemicircle 
AdG cultiiig XX in A and G. Join A V, G V, then 

V AG is the elevation required, and A(ZG is the half-plan 
of base. Xext divide AdG into any convenient number of 
parts, equal or unequal. The division here is into six, and 
the parts are equal ; to make them so being an advantage. 
Now let fall V V perpendicular to XX; and join V to the 
division points 6, c, d, e, f (to save multiplicity of lines this 
is only partially shown in the fig.). The lines TA, V6, 

Y c, &c., will be the lengths in plan of eeven lines from apes 



Fid. 3. 




to base of cone, that is, of seven generating lines. Now 
with V as centre and radii successively Y h, Y c,Y d,Y e, and 
V/, describe arcs respectively cutting X X in B, C, D, E, and 
F. Join V B, V C, &c. ; then as Y' V is the height of any 
elevation of either of the seven generating lines (see § 22a, 
p, 48, and § 55, p. 107) and as Y A, Y B, &c., are their 
plan lengths, we have in Y' A, Y' B,. Y' C, &c,, their true 
lengths. Y' G is not only so however, it is (§54) the shortesl; 
of all the generating lines of the cone. 

To draw the pattern of the cone with the Beam to corre^ 



no 



THE TINSxMlTHS' PATTERN MANUAL. 



spond with V G tho shortest generating line. Draw (Fig, 4) 
V' A equal to V A (Fig. 3), and with V as centre and 
V'B, V C, Y'D, V'E, V'P, and V G (Fig, 3) successively 
as radii describe arcs, respectively^ hb, cc, d d, ee, ff, and g g. 
"With A as centre and radius equal to A 6 (A & being the 
distance apart on the round of the cone at its base of the 
generating lines V A and V B') (Fig. 3) describe an arc 
cutting the arc 6 h right and left of V A in B and B. With 

Fia 4. 




these points B and B as centres and radius as befoie (the 
dietances apart of the extremities of the generating lines at 
the base of the cone being all equal), describe arcs cutting 
the arc cc right and left of V A in C and C. With same 
radius and the last-named points as centres describe arcs 
cutting d d right and left of V A in D and D. With 
D and D as ceatres and samo radiua describe area cutting e e 
right aad left of V' A in B and B, aad with B and B aa 



THE TINSMITHS' PATTERN MANUAL. 



Ill 



centres and same radius describe arcs cutting // right and 
left of V A in F and F. Similarly with same radius and 
F and F as centres find points G and G. Draw through A 
and the points B, C, D, E, F and G, right and left of V A 
an unbroken curved line. Also join G V right and loft of 
V A, then G A G V will be the pattern required. 

The dotted line^^ V B, V C, Y' D, &c., are not necessary for 
the solution of the problem, and are only drawn to show the 
position of the seven generating lines on the developed surface. 

(57.) As the plane of the elevation V A G bisects the cone 
(§§ 53 and 54), it is clear that the seven generating lines 
found, correspond with seven other generating lines on 
the other half of the cone ; so that in finding them for 
one half of the cone, we find thorn for the other, as the 
halves necessarily develop alike. 

(58.) Round articles of unequal slant or taper having 
their ends parallel are portions (frusta) of oblique cone^^ 




Definition. 

(59.) Frustum, — If an oblique cone be cut by a plane' 
parallel to its base, then . the part containirg the apes 
ifl still an oblique cone, as V A' U G' (Fig. 5), and the 



112 



THE TINSMITHS' PATTERN MANUAL. 



part A' A C G G' containing the base is a frustum of an 
oblique cone. A shaded representation of suol^ frustum 
will be found in Fig, 15. This frustum, however, 'differs 
from that of the right cone (§ 12, p. 33), in that, though 
having circular ends, its sides are not of equal, but of un- 
equal slant. Conversely, a round unequally tapering body, 
having its top and base parallel, is a frustum of an oblique 
cone. A tapering piece of pipe A' A G G' (Fig. 6) joining 
two cylindrical pieces which are not in line with each other 
is a fi-ustum of an bblique cone. 

(60.) Keferring to Fig. 5, it is evident that if the pattern 
for the larger con©, V A G, and the pattern for the smaller 
cone, V A' G', be drawn from a common centre V (Fig. 9), 
the figure G'GAGG'(Fig. 9) will be the pattern for the 
portion A G G' A' (Fig, 6) of the cone V A G. . The line 
C C shows a generating line of the frustum (6, p. 126). 

Fig. 7&. 



Fig. 7a 




(61.) A case that not unfrequently occurs 
here.. Suppose the diameters of the top and 



mention 
of a round 



THE TINSMITHS' PATTERN MANUAL. 113 

uneqnal-tapering body of parallel ends are very nearly equal, 
then it is evident that tlie apex of the oblique cone of which 
the body is a portion will be a long distance off, and, if the 
diameters be equal, then the body becomes what is called an 
oblique cylinder, (a cylinder is a round body without any 
taper at all). Of course this is an extreme case, but it is 
quite an admissible one. For, if the diameters of the ends of 
the body differ by only -j^ ^^^^j inch, then, clearly, however 
short the body may be, we are dealing with a frustum of an 
oblique cone, although so nearly a cylinder that, for almost 
any purpose occurring in practice, it could be treated as a 
cylinder. Later on the advantages of looking upon the 
oblique cylinder as a special case of frustum of an oblique 
cone, and considering its generating lines as parallel, will be 
seen. Such a frustum is represented in A A' G' G in Figs. 
la and 7&, the line 0' being the axis of the frustum and 
C C and F F' generating lines. ..A construction easily dealing 
with its development will be given presently. 



PROBLEM n. 

To draw the pattern of a round unequal-tapering "body with top 
and hase parallel {frustum, of an ohliquecone, as in Fig. 5), 
the diameters of top and hase, the height, and the inclination 
of the longest generating line leing given. 

First (Fig. 8) draw the elevation and half the plan of the 
body, thus : Draw a line X X', and at any point A • in 
it, make the angle X' A A' equal to the given inclination of 
the longest generating line. ' At a distance from X X' equal 
to the given height, draw a line A'G' parallel to XX'. 
From the poiirt A' where A'G' cuts A A', make A'G' equal 
to the diameter of the top, also make A G equal to the 
diameter of the base ; and join G G'. Then A A' G' G is the 
elevation of the body Now on A G describe a semicircle 
XdQ, this will be half the plan of the base. Produce A A', 
G G', to intersect in V ; this point will ba the elevation of 



114 



THE TINSMITHS' PATTERN MANUAL. 



tlie apex of the cone of which the body is a portion. Through 
"V' draw V V perpendicular to XX'; divide the semicircle 
into any convenient number of equal parts, here six, in the 



Fig. 8. 




points h, c, d, e, and/; vv-ith V as centre and ladii succos.sivcly 
V 6, V c, V (Z, &c., describe arcs respectively cutting X X in 
B, C, D, &c. ; and join B, C, D, &c=, to V by lines cutting A' G' 
in B', C, D', E', and F'. Then B B', C C, D D', &c., will be 
(see construction of last problem) the true lengths of various 
generating lines of the frustum (6, p. 126). 

Now to draw the pattern (Fig. 9) so that the seam shall 
correspond with G G' (Fig. 8) the shortest generating line. 
Draw (Fig. 9) V A equal to V A (Fig. 8) and with Y' 
(Fig. 9) as centre and radii successively equal to V B, V C, 
V' D, V E, V F, and V G (Fig. 8), describe, respectively, a.rcs 
L 6, cCydd, ee, ff, and a g. With A as centre and radius 
equal A 6 (Fig. 8) (see preceding problem for the reason of 
this), describe arcs cutting the arc hh right and left of VA 
in B and B. With these points, B and B as centres and 
radius as before, describe arcs cutting the arc cc right and 
left of VA in C and C. With same radius and tlio last- 



tlil-i TINSMITHS' PATTERN MANUAL. 



115 



named points as centres, describe arcs cutting dd riglit 
and left of V A in D and J). With D and D as centres and 
same radius, describe arcs cutting ee right and left of V A in 
E and E, and with E and E as centres and same radius 
describe arcs cutting // in F and F. Similarly, with same 
radius and F and F as centres find points G and G. Join the 
points B, C, D, E, &c., right and left of V A to V. With 
V as centre and V A' (Fig. 8) as radius describe an are 

Fig. 9. 




cutting V A in A'. With same centre and V'B' (Fig. 8) as 
radius describe an arc h'h' cutting V B right and left of V' A 
in B'. With same centre and V'C (Fig. 8) as radius 
describe an arc c' c' cutting V'C right and left of V A in 0'. 
Similarly, with the same centra and V D', V E', V F', and 
V G' (Fig. 8) successively as radii describe, respectively, arcs 
d'd', e'e\ff', and g'g', cutting V'D, V E, V F, and V'G, 
right and left of V'A respectively in D', E', F', and G'. 
Draw through A arid the points B, C, D, E, F, and G, right 



lis THE TliJSMITHS' PATTERN MANUAL. 

and left of V A an tinbroten cnrved line. Also di'aw tlirougli 
A' and tlie points B', C, D', E', F, and G', right and left of 
V A an nnbroken curved line. Then G A G G' A' G' will 
be tbe pattern required. 

The small semicircle (Fig. 8) and ^perpendicular lines from 
its extremities are not needed in tliis problem, but are 
introduced in illustration of that next following. 

(62.) In the appKcations of. i ho oblique cone, it is generally 
in the form next following that the problem presents itself. 



PEOBLEM m. 

To dravj the pattern of a round unequal-tapenng body with top 
and base parallel (/rusium of oblique cone), its plan and the 
perpendicular distance between tlie top and base {the height of 
the body) being given. 

The working of this problem should be carefully noted, 
for the reason just above stated. 

Let aga' g' (Fig. 10) be the given plan of the body. The 
side lines of the plan are jiot drawn, but only the circles of 
its top and base, as we do not make use of the side lines. In 
fact, all that wo really make use of is, as will be presently 
seen, the halves of the plan-circles of the top and base of the 
body, not, however, placed anyhow, but in their proper positions 
relatively to one another in plan. This is the all-essential point. 
If the plan-circle a' g' (Fig. 10) of the top of the frustum 
were further removed than represented from the pi an -circle 
a {7 of its base, we should, by the end of our working, get at 
the pattern of some other oblique-cone frustum than that the^ 
pattern of which we require. Through 0' the centres of 
the circles draw an indefinite line a V, and draw any line 
XX parallel to 00'. Through a and g draw a A and gG 
perpendicular to X X, and through a' and g' draw indefinite 
lines a' A" and g' G' perpeudicular to X X, and cutting it in 



THE TINSMITHS' PATTERN MANUAL. 



Ill 



points A', G'. Mate A' A" and G' G" each equal to the ^ven 
height or perpendicular distance tliat the top and base are 
apart. Join A A" and G G", and produce these lines to inter- 
sect in v. The problem can now be completed by Problem II. 

Fio. 10. 




It should be noted that it is only necessary to draw half the 
plans of the top and base as shown by the semicircles A G 
and A' G' in Fig. 8. 



The case of large work, whero long radii would be incon- 
venient, we treat as a sepg^rate problem, and we will suppose 
the dimensions given to be those of Rroblem 11. The 
method is also suitable where there is little difference 
between the diameters of top and base of the body. 



118 



THE TINSMITHS PATTERN MAJfUAL. 



PROBLEM IV. 

To draw without long radii the pattern of a round unequal- 
tapering hody with top and "base parallel, the diameters of the 
top and hose, the height, and the inclination of the lonnest 
generating line being given. 

Draw any line X X' (Fig. 11) and at point A in it make 
the angle X' A A" equal te the given inclination of the 
longest generating line. At a distance from X X' equal to 
the given height draw a line A"F parallel to XX'. Make 
A E equal to the longer of the given diameters, and on it 
describe the semicirele A c E. From the point A" where 

Fia. II. 




A" F cuts A A" let fall A" A' perpendicular to X X. Make 
A' E' equal to the shorter of the given diameters, and on it 
describe the semicircle A' c' E'. Next find c and c' the mid- 
points of the semicircular arcs. These are easily found by 
drawing lines fnot shown in the fig.) through and 0' (the 



THE trNSMITHS- PATTERN M-VNl'AL. 119 

centres from wluch tLe semicircles are described) perpen- 
dicular to X X' and cutting the eemicircular arcs in the 
required points c and c\ Join c c' and produce it meeting 
X X' in V ; tnis point will be the plan of the apes of the cone 
of which our tapering body is a frustum. This point is by 
no means always within what may be termed workable reach, 
as for instance where the two semicircles are nearly equal 
the lines X X' and ec' therefore very nearly parallel, and the 
producing c c' to V impracticable. We will work under both 
suppositions. 

(63.) If V is accessible, then divide the larger semicircle 
into any convenient number of parts (four parts only are 
taken in the figure in order to keep it clear), as Ah,bc, c d, 
d E. Join h and dioY (c iig already thus joined, the lines 
from 6 and <J to V are only drawn in the fig. as far as the 
smaller semioircle) by lines cutting the smaller semicircle 
in 6' and d'. Then A A', h b', c c\ &c., are the plans of gene- 
rating lines of the frustum (tapering body), and in order to 
draw its pattern their true lengths must be found. 

(64.) If V is inaccessible, then dieide the smaller semi- 
circle into four parts (the same ' convenient number ' of parts 
that the larger semicircle was divided into), in the points h', 
c\ and <?', and join 6 \>\ c c', and d d\ 

The true lengths of h h\ c c\ &c., are found as follows : From 
E' draw a line perpendicular to X X' and cutting A" F in F, 
and join EF. Then E F is the true length of EE'. Now 
make E' I) equal tp d d' and join D F ; then D F is the true 
length of dd'. Next set off E' C equal to c c', and E' B equal 
to 5 &' ; and join C F and B P. Then C F and B F are the 
true lengths respectively of cc' and 6 6'. The true length of 
A A' we already have in A A", and as this is the longest 
generating line of the frustum, E F will be the shortest. 

We proceed now to find the distance the points A and 6', 
6 and c', c and d', &c., are apart, which we do by finding the 
true lengths of the lines A 6', 6 c', c d', and d E', joining 
the points. Through 6' draw 6' 6" perpendicular to A 6', and 
eyual to tho given height,- Join A 6" ; then A b" may be 



120 



THE TINSMITHS' PATTERN MANUAL. 



taten as the true lengtli of A h\ Similarly, Hirongli c', (t, 
and E' draw lines equal to the given height, and perpen- 
dicular to h c', c d', and d E' respectively. Join h c", c d", and 
de" ; then h c", cd", and de" may be taken as th'' true lengths 
required. ■ 

To diav? the pattern (Fig. 12) the seam to correspond with 
the shortest generating line. Draw A A' equal to A A" 
(Fig. 11) and with A and A' as centres and radii respectively^ 
Ah" and A' 6' (Fig. 11) describe arcs intersecting in &'. 
Next with 6' and A as centres and radii respectively B F and 
A h (Fig. 11) describe arcs intersecting in 6. Then A, 6, A', 6', 
aro points in the curves of the pattex"n. With h and 6' as 




centres and radii respectively he" and V c' (Fig. 11) describe 
arcs intersecting in c\ With c' and 6 as centres and radii 
respectively C F and b c (Fig. 11) describe arcs intersecting 
in c; and with c and c' as centres and radii respectively c d" 
andc'd'(Fig. 11) describe arcs intersecting in d'. With d' 
and c as centres and radii respectively D F and cd (Fig. 11) 
describe arcs intersecting in d. Similarly find E' and E. 
Draw unbroken curved lines through A & c d £ and A* V d d' E 



THE TINSMITHS' PATTERN MANUAL. 



121 



and join EE'; iliat will give us Lalf tlie pattern. By like 
procedure we find the otlier half of the pattern, that to the 
left of A A'. 

(65.) The lines 6 h\ c c\ &c., and the dotted lines' A h\ h c', &c., 
are drawn in Fig. 12 simply to show the position that the 
lines which correspond to them in Fig. 11 {bh\ A 6', &c.) 
take upon the developed surface of the tapering body. It is 
evident that it is not a necessity to make distinct operations 
of the two halves of the pattern ; for as the points 6', 6, c\ c, 
&c., are successively found, the points on th'^e left of A A' 
corresponding to them can be set off. 



PROBLEM V. 

To draw Oie pattern of an oblique cylinder {inclined circular pipe 
for example), the length and inclination of the axis and the 
diameter being given. 

Draw (Fig. 13) any line A' G', and at the point A' in it 



FiQ. 13, 



A' B" C" D' £' F 




make the angle G' A' A" equal to the given inclination of the 
axis. Make A' G' equal to the given diameter, and draw a 
line G' G" parallel to A' A". Make A' A" and G' G" each equal 



122 



THE TINSMITHS' PATTERN MANUAL. 



to the lengtli of the cylinder (the length of a cylinder is the 
length of its axis) and join A" G". Then A' A" G' G'! is the 
elevation of the cylinder. Now on A' G' describe* a semi- 
circle, and divide it into any number of equal parts, in the 
points 1, 2, 3, &c. : through each point draw lines perpendi- 
cular to A' G', meeting it" in points B', C',D',&c., and through 
B', C, D', &c., draw lines parallel to A' A". Draw any line 
A G perpendicular to A' A" and G' G", cutting the lines B' B", 
C C", &c., in points B, G, &c. Next makg B b equal 
to B'l, Gc equal to C'2, Bd equal to D'3, Ee equal 
to E' 4, and F/ equal to F' 6, and draw a curve from 
A through the points h, c, d, &c., to G. It is necessary 
to remark that this curve is not a semicircle, but a 
semi-ellipso (half an ellipse). 




To dra^ the pattern (Fig. 14). Draw any line A A, 
aud at about its centre draw any line G" G' perpendicular tp 



THE TINSMITHS' PATTEUN MAXUAL. 123 

it and cutting it in G. From G, right and left of it, 
on the line A A mark distances G F, F E, E D, D C, C B, and 
BA equal respectively to the distances Gf,fe, cd, d c, 
&c. (Fig. 13). Through the points F, E, D, &c., right 
and left of G, draw lines parallel to G" G'. Make G G', g'^G" 
equal to G G', G G" (Fig. 13) respectively. Similarly 
make F F', F F", E E', E E", D D', D D", &c., right and left of 
G' G" equal respectively to F F', F F", E E', E E", D D', D D" 
&c. (Fig. 13). Draw an unhroken curved lino from G" 
through F", E", D", &c., right and left of G" and an unbroken 
curved line through F', E', D', &c., right and left of G\ The 
€gure A"G"A"A'G'A' will be the pattern required. The 
two parts G"A"A'G' of the pattern are alike in every 
tespect. 



124 THE TINyMITIIS' PATTERN MAISUAL. 



CHAPTER II. 

IJNEQUAL-TAPERINa BoDIES, OF WHICH ToP AND BaSE ARE 

Parallel, and their Plans. 

(66.) Before going into problems showing how to draw tho 
patterns of unequal-tapering bodies with parallel ends, 
bodies which are (^as the student will realise as he proceeds) 
partly or wholly portions of oblique cones, it will be neces- 
sary to enter into considerations in respect of the plans of 
frusta of such, cones (see § 58, p. Ill), similar to those 
appertainiEg to the plans of frusta of right cones treated 
of in Chap. V., Book I. ; but to us of greater importance, 
because the constructions in problems for the setting out 
of patterns of bodies having unequal taper or inclination 
of slant are a little more difficult than those in problems for 
patterns of equal-tapering bodies. The chapter referred to 
may be now again read with advantage. 

As much use will be hereafter made of the terms Pro- 
poriionale Arcs and Similar Area we now define them. We 
also extend our explanation of Corresponding Points, 

Definitioks. 

(67.) Phoportional Arcs: Similar Arcs.— Arcs are pro- 
portional when they are equal portions of the circumferences 
of the circles of which they are respectively parts; they 
are similar when they are contained between the same 
generating lines. Similar arcs are necessarily proportional. 
In Fig. 15 the arcs A D and A' D' are proportional because 
each is a quarter of the circumference of the circle to which 
it belongs. They are similar because the generating lines 
V A and V D contain them both. 

(68.) CoRRESPONDiNa Points.— Points on the same gene- 
rating line are corresponding points (compare § 24) ; thus, 

r 



THE TINf^MlTUS' PATTERN MANUAL. 



125 



'-he points A and A are corresj.onding points, because they 
are on the same generating line V A ; also the points C and 
C OQ the generating line V C The point A' on V A is the 



Fig. 15, 




plan of A on V A ; the point C on V C is the plan of C on 
V ; and so on. 

(69.) From the figure, which shows a frustum of an oblique 
cone standing on a horizontal plane, it will be seen that the 
plan of a round unequal-tapering body (frustum of oblique 
cone) consists mainly of two circles G A D B, C A' D' B', the 
plans of the ends of the body. In Fig. 16 is shown a 
complete plan of an oblique-cone frustum. With the con- 
necting lines of the two sides we are not concerned, but may 
simply mention that they are tangents (lines which touch 
but do not cut) to the circles. Further from Fig. 15 it will 
bo seen that, completing the cone of which the tapering 
body is a portipn. 

a. The plan of the axis (line joining the centres of the 
ends) of a frustum is a line joining the centres of the circles 
which are the plans of its ends ; thus, the line O 0' is the 
plan of the axis 0" (s§e alsp 0', Fig. 16). 



126 THE TINSMITHS' PATTERN MANUAL. 

Similarly, the plan of tlie axis of the complete cone, is 
plan, produced, of tlia axis of its frastum; thus, O Y is the 
plan of O v. 

6, The plan, produced both ways, of the axis of a frustum 
contains the plans of the lines of greatest and least inclina- 
tion on the frustum (see § 52) ; that is to say, of the longest 
and shortest lines on it. Thus, O O', produced both ways, 
contains the plans of A A' and B B'. It is convenient to 
regard lines joining corresponding points of a frustum 
(corresponding points of a frustum are points on one and 
the same generating line of the complete cone) as generating 
lines of the frustum. (See in connection with this, § 46 j. 
Then lines A A', B B', fur instance, may bo spoken of as 
generating lines of the frustum represented. 

Similarly, the plans of the longest and shortest generating 
lines of a cone are contained in the plan, produced, of the 
axis of its frustum; thus, the plans of V A and Y' B are 
contained in O 0' produced, both ways. 

c. The line, produced, which joins the centres of the plans 
of the ends of a .frustum, contains the plan of the apex of 
the cone ; thus, O O', produced, contains Y the plan of Y'. 

d. The line, produced, which joins the plans of corre- 
sponding points of a frustum (see definition, §68) contains 
the plan of the apex of the complete cone, and, produced only 
as far as the plan of the apex is the plan of a generating 
line of the cone; thus, C and C being corresponding -points 
on the cone, the plan, C C, produced, of the joining C and C, 
contains Y ; and C Y is the plan of the generating lino 
CY'. 

e. The plans produced of ali generating lines of a frustum 
intersect the plan of its axis produced, in one point, and that 
point is the plan of the apex of the ,completo cono; for 
example, the plans produced, of the generating linos C C 
and D D' of the frustum intersect 0' produced in Y. 

(70.) It follows from e that the plan of the apex of the 
complete cone of which a given frustum is a portion can 
easily be found if we have given the plans of the ends of the 



THE TINSMITHS' PATTERN MANUAL. 127 

frustum and tlie plans of two corresponding points not in 
the line passing through the centres of the plans of its ends. 
This is a valuable fact for us, as it spares us elevation: 
drawing which in many cases is very troublesome, and 
•indeed, sometimes practically impossible, as, for instance, 
where an unequal- tapering body is frustum of an exceed- 
ingly high cone the axis of which is biit little out of the 
perpendicular. This is a case in which although the apex 
caniiot be found in elevation because of the great length of 
the necessary lines, it can readily be found in plan, because, 
in plan, the requisite lines are short. An oblique cone may 
of course not only be exceedingly long, but also very greatly 
out of the perpendicular. In this case it is impracticable 
anyhow to find the plan of the apex. Problem IV., just 
solved, meets both casus. It was by e that we there foun'd 
the plan of the apex wlien accessible, that is, where the lines 
of the plan are not unduly long (see Fig. 11) by joining the 
plans of corresponding points c and c' (c and c' are corre- 
sponding points in that they are mid-points on the balf- 
plans of the ends of the frustum, and therefore necessarily 
on one and the same generating line), and producing c c' to 
intersect 0', the line joining the centres of the plans of 
tbe ends, that is to intersect the plan produced of the axis 
of the frustum. 

(71.) Passing the foregoing tinder review, it will be seen 
that if we have two circles which are the plans of the ends 
of a round unequal-tapering body (frustum of an oblique 
cone) standing on a horizontal plane, and the circles are in 
their proper relative positions as part plan of the frustum, 
then the line produced, one or both ways as may be necessary, 
which joins the centres of the circles, contains : — 

The Plan of the Axis of the frustum (see a, p. 125). 

The PiJLN of the Axis of the cone of which the frustum is 
a part (see a, p. 125). 

Tlie Plan of the Apicx of the cone (see c, p. 126). 

The Plans of the Longest and Shortest Gknerating lines of 
the frustuia (see h, p. 126). 



128 THE TINSMITHS' PATTERN MANUAL. 

The .Plans of tlie Longest and Shortest Generating 
Lines of the cone, of which the fiustum is a part ("see &, 
p. 126). 

The Plans of the Lines of Greatest and Least Inclination 
of the frustum (see 5, p. 126). 

The Plans of the Lines of Greatest and Least Inclina- 
tion of the cone, of which the frustum is a part (see 
§ 52, p. 107). 

And this is a matter of very great practical importance, as 
will be seen later on. 

(72.) As with circles "under the conditions stated, so exactly 
with arcs which form the plans either of the ends or of 
portions of the ends, of an unequal -tapering body. 

(73.) Further, referring to c and d of p. 55 as to round 
equal- tapering bodies we are now in a position to deduce (see 
Fig. 15) the following as to round unequal-tapering bodies. 

/. The plan of a round unequal-tapering body with top 
and base parallel (frustum of oblique cone) consists essentially 
of two circles, not concentric, definitely situate relatively to 
onQ another. See Fig. 16. 

Fig. 16. 




Similarly the plan of a portion of such round unequal- 
tapering body (frustum of oblique cone) consists essentially 
of two ai'cis definitely situate xelatively to one another, and 



THE TINSMITHS' PATTERN MANUAL. 129 

not concentric. See Fig. 17. 0' is the axis of the complete 
frustum. 

g. Conversely. — If two circles, not having the same centre, 
definitely situate relatively to one another, form essentially 
the plan of a tapering hody having parallel ends, that body 
is a round unequal-tapering body (frustum of oblique cone). 

In Fig. 16, if the two circles represent essentially the 
plan of a tapering body having parallel ends, then the body, 
of which the circles are the essential plan, is a round uneoual- 
body (frustum of oblique cone). 

Similarly if two arcs definitely situate relatively to one 
another, and not having a common centre, form the essential 
part of the plan either of a tapering body or of a portion of 

Fig. 17. 




a tapering body having parallel ends, then that body or por 
tion is a portion of a round unequal-tapering body (frustum 
of oblique cone). 

\ In Fig. 17 if the arcs form the essential part of the plan, 
either of a tapering body or of a portion of a tapering body 
having parallel ends ; then the body or portion of body, of 
which that fig. is the plan, is a portion of a round unequal- 
tapering body (frustum of oblique cone). In the particular 
plan represented the arcs are similar; the points B and 
B', and A and A' are therefore corresponding points. 



13a 



THE TINSMITHS' PATTERN MANUAL. 



We will now proceed to draw the plans of some imequal- 
tapering bodies, of which patterns will be presently set out. 



PEOBLEM VI. 

To draw the plan of an unequal-tapering body with top and hase 
parallel and having straight sides and semicircular ends 
(an " equal-end " hath with semicircular ends), from given 
dimensions of top and bottom. 

Draw A'B'C'D' (Fig; 18) the plan of the bottom by 
Problem XVI., p. 20. Bisect A'B' in 0, and through O 
draw C D perpendicular to A' B'. Make A and B each 

Fig. 18. 




equal to half the length, and and D each equal to half 
the width of the top. The plan of the top can now be drav/n 
in the same manner as that Qf the bottom, completing the 
plan required. 



- TOm xx^N^Mliiift vkli 'E-RT^ 'HK^TVAL. 



131 



PROBLEM VII. 

To draw tlie plan of an oval unequal-tapering body with, top and 
base parallel (aw oval bath), from given dimensions of top 
and bottom. 

Draw (Fig. 19) A'B'C'D' the plan of tlie bottom, llie 
given length and width, by Problem XII., p. 13 ; and make 

Fio. 19. 




O A and B eacli equal to half the length, and C and D. 
each equal to half the width of the top. The plan of the top 
can now be drawn in the same manner as that of the bottom ; 
this completes the plan required. 



132 



\THE TINSMITHS' PATTE:5N MANUAL? 



PEOBLEM VIII. 

To draw tne plan of a tapering body with top and base parallel 
and having oblong bottom icith se^nicircular ends and 
circular top [tea-bottle top), from given dimensions of top and 
bottom. 

Draw (Fig. 20) the plan of the oblong bottom by 
Problem XVI., p. 20, and with O tbe intersection of the axes 
of the oblong as centre and half the diameter of the top as 
radius, describe a circle. This completes the plan. 

Fig. 20. 




We here, for the first time, extend the use of the word 
'axes' (see Problems XII. and XIV., pp. 13 and 15). It is 
I convenient to do so, and the_meaning is obvious. 



PEOBLEM IX. 

To draw the ^lan of a tapering hody with top and base parallel, the 
top being circular and the bottom oval (^oval canister-top), 
from given dimensions of top and bottom. 

Draw (Fig. 21) the plan of the oval bottom by Problem 
KII., p. 13, and with the intersection of the axes as centre. 



*rHE TINSMITHS' PATTERN MANUAL.^ 



133 



and half tlie given diameter of the top as radius, describe a 
circle. This completes the plan. 



Fig. 21, 




I PEOBLEM X. 

To draw the plan of a tapering body with top and base parallel 

and having oblong base with round corners and circular tov^ 

from given dimensions of the top and bottom. 

Draw C^ig. 22) %Ue plan of the oblong bottom by Problem 

XV., p; 19, and with the intersection of the axes of the 

Fig. 22. 




oblorg bottom as centre and half the diameter of thie top as 
radius, describe a circle. - This completes the pl^m. - 



134' THE TINSM-ITllS' PATTERN MAXUAL. 

FEOBLEM XI. 

To draw the jplan of an (hiford hip-bath. 

Fig. 23 is a side elevation of the bath, drawn here only to 
make the problera clearer, not because it is necessary for the 
working. No method that involves the drawing of a full- 
size side elevation is pr-actical, on acco.unt of the amount of 
space that would be required. 

Via. 23. 




The "bottom of an Oxford hip-bath is an egg-shaped oval. 
The portion O X' of the top is parallel to the bottom A' B', 
and the whole XX' top, the portion O X E of the bath being 
removed, is also an egg-shaped oval. In speaking of the 
plan of the ' bath,' we mean the plan of the ~ X X' B' A' 
portion of it, as the plan of this portion is all that is necessary 
to enable us to get at the pattern of the bath. 

We will first suppose the following dimensions given : — 
The length and width of the bottom, and the length of the 
X X' top, the height of the bath in front, and the inclination- 
of the slant at back. 

First draw (Fig. 25) the plan of the bottom A'D' B' C by 
Froblem XITI., p. 14. To draw the horizontal projection of . 



THE TiNSMItllS' PATTERN MANUAL. 



135 



the X X* top, make (Fig. 24a) the angle A A' E equal to the 
given inclination of the slant at the hack. Through A' draw 
A' H perpendicular to A A', and equal to the height of the 
bath in front ; through H draw H X parallel to A A' and 
cutting A'E in X; and draw XA perpendicular to XH; 
then A A' will be the distance, in plan, at the back, between 
the curve of the bottom and the curve of the X X' top 
(Fig. 23). Make A' A (Fig. 25) equal to A A' (Fig. 24a), and 
make A B equal to the length of the X X' top. With as 
centre and A as radius describe a semicircle ; the remainder 
of the oval of the X X' top can now be drawn as was that of 
the bottom. This completes, as stated above, all that is 
necessary of the plan of the bath to enable its pattern to be 
drawn. 



Fig. 24a, 



Fig. 246. 




If the width of the X X' top is given, and not fno inclina- 
tion of the bath at back, make O A (Fig. 25) equal to half 
that width, and proceed as before. The seam in an Oxford 
hip-bath, at the sides, is on the lines of which C C and 
D D' are the plans. 

If tha length of the X X' top (Fig. 23) is not given, it can 
be determined ii. the follov/iug manner : — 

Let ^he angle X B' B (Fig. 246) represent the inclination 
of tl\» t;lant of the front, and B' X its length. Through X 



1.1^5 



THE TIXSMITIIS- PATTEKN M\NUAL 



draw XB perpendicular to S'B; tlien BE' will be the dis- 
tance in plan, at tlie front, betweca. the ciirve of the bottom 
and the curve of the XX' top (Fig. 23), and this distance, 
marked from B' to B (Fig. 25), together with the distance 
A' A at the back end of the plan of the bottom, fixes the length 
required. 

Fig, 25. 




If the lengths only of the slants of the bath at back and 
front are given and not their inclinations, the plan of the 
X X' top can be drawn as follows : — 

Draw two lines X B, B' B (Fig. 246) perpendicular to one 
another and meeting in B ; make B X equal to the given 
height of the bath in front, and with X as centre and radius 
equal to the given length of the slant at the front,-* describe 
an arc cutting B' B in B'. Make B'B (Fig. 25) equal to 
BB' (Fig. 246), and B A equal to the given length of the 
X X' top ; this will give the distance A' A. Now make A A' 
(Fig. 24a) equal to A A' (Fig. 25), dra'c^ AX and A'H per- 
pendicular to A A' and equal to the given height of the bath 
in front ; join H X and draw A' E, through X, equal to the 
length of the slant~at the back ; the remainder of the plan 
of the X X' top can then be drawn as already described. 
, Jt will be useful to show here in this problem how to 



THE TINSMITHS' PATTERN MANUAl.. 137 

complete the back portion already comtnenced in Fig 2 la 
of the side elevation of the bath. Make A' E equal to the 
slant (§ 4, p. 24) at back, which must of course be given, 
and make XO equal to half the given width of the XX' 
top ; join E, and draw O O perpendicular to A A' produced ; 
then A',E is the elevation reouired. 



PEOBLEM XII. 
To draw the plan of an Athenian hip-hath or of a sitz-hath. 

Fig. 26 is a side elevation of the bath, drawn for the reason 
mentioned in the preceding problem. 

Fig. 26. 





The bottom of an Athenian hip-bath or a sitz-bath is an| 
ordinary oval. The portion X' P of the top is parallel to the 
bottom A' B', and the whole X X' top, the portion F X E 
of the bath being removed, is also an ordinary oval. Simi- 
larly as with the bath of the last problem ; we mean by plan 
of the ' bath,' the plan of the XX B' A' portion of it ; nt 
more being required for the drawing of the pattern of the 
bath. 

Vy© will first suppose the given dimensions to be those of 



138 THE TiNSMItiiS' iPATTERN MANUAL. 

the bottom and tlie X X' top of the bath^also height of the 
bath in front. 

First draw A' D' B' C (Fig, 27) the plan of the bottom by 
Problem XII., p. 13. To draw the plan of the XX' top 
(Fig. 26) set off O A and B each equal to half the given 
length of that top, and and OD each equal to half its 
given width. The plan of the X X' top can now be drawt^ 

Fig. 27. 




as was that of the bottom. This completes, as stated aV^ve, 
all that is necessary of the plan of the bath to enabU its 
pattern to be drawn. 

If the length of the X X' top (Fig. 26) is not given but the 
inclination of the slant at front and back, these inclinations 
being the same, the required length can be determined iii the 
following manner : — 

Make the angle AA'E (Fig. 28a) equal to the given 
inclination. Through A' draw A' H perpendicular to A A' 
and equal to the given height of the bath in front ; through 
H draw H X parallel to A A' and cutting A' E in X, and 
draw X A perpendicular to A A' ; then A A' will be the 
distance in plan, at back and front, between the curve .of the 



Till!; TINSMITHS' PATTERN MANUAL. 



139 



bofetom and the curve of the X X' top. Make A A' (Fig. 27) 
and B E' each equal to A A (Fig. 28a) ; then A B will be the 
length required. 



Fig. 28a 



Fig. 28&. 





If the length of the X X' top of the bath (Fig, 26) is not 
given, nor the inclination of the slant at front and back, but 
only the length of the slant at front, the required length can 
be thus ascertained. 

Draw two lines XB, B' B (Fig. 286) perpendicular to one 
another and meeting in B ; make B X equal to the given 
height of the bath in front, and with X as centre, and radius 
equal to the length of the slant at the front, describe an arc 
cutting B B' in B'. Make A' A and B' B (Fig. 27) each equal 
to B B' (Fig. 286), then A B is the length wanted. The 
remainder of the plan can be drawn as described above. 

By a little addition to Fig. 28a we get at the back portion 
of the side elevation of the b'.th. It will be usrful to do 
this. Produce A'X and make A'E equal to the slant at 
back, which must of course be given. Then, on the plan 
(Fig. 27), E being the meeting point of the end and side 
curves of the oval A D B C, draw E F perpendicular to A B. 
Make X F (Fig. 28a) equal to AF (Fig. 27) ; join F E ; this 
completes the elevation required. 



140 



THE TINSMITIJS' PATTERN MANUAL, 



PEOBLEM XIII. 

To draid the plan of an oblong taper oath, the she of (he top and 
boilom, the height, and the slant at the head being given. 

To draw D E F C (Fig. 30) tlie plan of the top. Draw A B 
equal to tlie given length of the top, and through A and B 
draw lines perpendicular to A B. Make A E and A D each 
eqnal <o half the width of the top at the head of the hath, 
and BF and B C each equal to half the width of the top at 
the toe ; and join E F and D C. Next from E mark off along 
E P and E D equal distances, E G and E H, according to the 
size of the round corner required at the head. (It will be 
xiseful practice for the student to work this problem, com- 
mencing with the plan of the bottom, and its smaller corners, 
for the reason given in § 27a, p. G3). Through G and H draw 



Fki. 29a. 





lines perpendicular to E F and E D respectively, intersecting 
in O ; and with O as centre and O G as radius describe an 
arc H G to form the corner. The round corners at D F C, 
&c., are drawn in like manner. 

To diaw the plan of the bottom. . Lot the angle A" A' A 
(Fig. 29a) be the angle of the inclination of the slant at the 
head, and A' A" the length of the slant. Through A" draw 



THE TINSMITHS- PATTERN .MANUAL. 



ui 



A." A perpendicular to A A', then A A' will lie the distance 
between the lines, in plan, of the top and bottom at the 
head. Make A A' (Fio:. 30) equal to A A' (Fig. 29rt), and 
A'B' equal to the length of the bottom. Through A' and 
B' draw lines each perpendicular to A B ; make A'E' and 
A' D' each eqxial to half the width of the bottom at the head, 
and B' F' and B' C each equal to half the width of the 
bottom at the toe. Join E' F' and D' C. The round corner 
of the bottom at the head must be drawn in proportion to 
the round corner of the top at the head, and this is done in 
the following manner. JoinEE' and produce it, to meet 

Fig. 30. 




A B in P, and join H P by a line cutting D' E' in H' ; mako 
E' G' equ;il to E'H', and complete the corner from centre O' 
obtained as was the centre O. Draw the other corners in 
similar way, and this will complete the plan required. The 
D corner is like the E corner ; the corners also at F and 
correspond. Similarly with the E' and D', and F' and C 
corners. 

If the length of the bath is given and the length of slant at 
(but not its inclination) head or toe^ the diistance A A' can be 



142 THE TINSMITHS' PATTERN _MANUAL. , 

found by drawing two lines A" A, A' A (Fig. 29a) perpendicular 
to one another and meeting in A, and making A A" equal to the 
given height ; then, with A" as eentr© and A" A', the given 
length of the slant at th© head, as, radius, deserib© an axe 
cutting A A' in A'. Then A A' is the distance rsc^iaired. 
Similarly fFig. 296) the distance B B' can he found. 



THE TINSMITHS' PATTBUN MANUAL. 



143 



CHAPTER III 

Patterns foe Aktioles of Unequal Taper or Inclination 
OF Slant, and having Flat (Plane) Surfaces. 

(Cl-ASS IL Suhdivision 5.) 

Articles of unequal taper or inclination of slant, and having 
plane or flat surfaces (hoppers, hoods, &c.), axe frequently 
portions {frusta) of oblique pyramids, or parts of such 
frusta. 

Definitions. 

(74.) Oblique Pyraix^id : Frnstum of Oblique Pyramid :— 
Oblique pyramids have not yet been defined. For our 
purpose it will be sufficient to define an oblique pyramid 
negatively, that is, as a pyramid which is not a right 
pyramid ; and when cut by a piano parallel to its baso (that 

Fig. 31. 




is, when truncated), to define its frustum (§ 33, p. 69) as the 
frustum of a pyramid which is not a right pyramid. In the 
oblique pyramid the facGS are not all equally inclined,. 
Articles of which the faces are not all equally inclined are 



144 



THE TINSMITHS' PATTERN MANUAL. 



not nei^essarily portions of oblique pyramids. One such 
case will be given later on. The problems immediately 
following deal with articles of which the faces are not all 
equally inclined, but which are portions of oblique pyramids. 
(75.) Further, an oblique pyramid, when it has a base 
through the angular points of which a circle can be drawn, 
can bo inscribtd in an ob.ique cone like as a right pyramid 
in a right cone, and this prope ty gives constructions fur 
solving most of our oblique-pyramid problems, somewhat 
similar to those in Book I., Chapter VI., where the right 
pyramid is concerned. Fig. 31 represents an oblique 
hexagonal pyramid inscribed in an oblique cone. This fig. 
should be compared with Fig. 31, p. 67, The edges of the 
oblique pyramid are generating lines of the cone. 

Fiff. 32. 




(76.) Also from Fig. 82 it will be seen that the plan of a 
frustum of an oblique hexagonal pyramid standing on a hori- 
zontal plane consists of two hexagons A E D and A' B' 0' D' 
(the plans of the eads), whose similarly situated sides, A B aad 



THIi; TINSMITHS' PATTERN MANUAL. 145 

A' B', B C and B' C, C D and C D' for instance, are parallel, and 
whose corresponding points (^ GS, p. 124) A, A' and B, B', for 
instance, are joined by lines A A', B B', which are the plans of 
the edges of the frustum. Just as in the case of the frustum 
of the oblique cone (see d and e, p. 126), if a line joining 
corre-ponding points in plan be produced, it will contain the 
2)lau of the apex of the complete pyramid of which the 
'frustum is a portion ; and if another such line be produced 
to intersect the first line, the point of intersection will be 
that plan of apex. For example, the lines A A', B B' and 
C C produced meet in a point which is the plan of the apex 
of the pyramid of which the frustum ABDD"B"A" is a 
portion. 

(77.) From this it follows that if the plan of a tapering body 
with top and base parallel and having plane or flat surfaces 
be ^given, we can at once determine whether the tapering 
body is or not a frustum of an oblique pyramid by producing 
the plans of the edges. If these meet in one point, then the 
given plan is that of a frustum of an oblique pyramid. 



TKOBLEM XIV. 

To draw the jjattern of an oblique pyramid. 
Case 1. — Given the plan of the pyramid and its height. 

LetABCDEFV (Fig. 33) be the plan of the pyramid 
(here a hexagonal pyramid). V being the apex, and O V the 
plan of the axis. Draw X X parallel to O V, and through 

V draw V V jierpeudicular to X X. and cutting it in v ; make 

V V equal to the given height of the pyramid. Next mako 
V a, vb, vc, V d; v e, and vf equal respectively to Y A, V B, 
VC, VD, VE, and V F, tiie plans of the edges of the 
pyramid. Joining V'a, V'65 Vc, &c., will give the tr\ie 
lengths of these edges. 

To draw the pattern of the pyramid with tlie seam at the 
edge V A. Draw V A (Fig. 34) equal to V a TFig. 33) ; with 



116 



TllR T^N^Mrll[S' PAT'J EEX MANUAL. 



V as centre atkd V h (Fig. 33) as radius describe arc h, and with 
A as centre and A B (Fig. 33) as radius describe an arc inter- 
secting the arc b in B. The other points 0, D E, F, A, are 

Fig. 33. 




THE TINSMITUS' PATTERN MANUAL. 



U1 



found in similar manner. Thus, witli V (Fig. 34) as centre 
and V'o, Y' d. Ye, V'/, and Y' a (Fig. 33) successively as 
radii, describe arcs c, d, e, f, and a (Fig. 34). Next, with 
B (Fig. 34) as centre and BC (Fig. 33) as radius, describe an 
arc inteiBecting arc c in C ; with D (Fig. 33) as radius and 
C (Fig. 34) as centre describe an arc cutting arc d in D ; with 
D (Fig. 34) as centre and D E (Fig. 33) as radius describe art 
o,rc intersecting arc e in E ; and so on for points F and A. 
Join A B, B C, C D, DE, E F, F A and A V, and this will 
complete the pattern required. 




Joining the points B, C, D, ^-c, to V, it will "be seen that 
the pattern is made up of a number of triangles, each triangle 
being of the shape of a face of the pyramid, also that the 
construction of the pattern is very similar to the constmction 
of that of an oblique cone. 

Should it be inconvenient to draw X X in the position 
ishowu in Fig. 33^ the true lengths of the edges of the pyramid 



148 The tinsmiths- pattern ManuaL 

can be found in tlio following manner. Draw X X quite 
apart from the plan of the pyramid, and from any point v in 
it draw v V perpendicnlar to X X,. and equal to the height of 
the pyramid, and proceed as just described. 

, Case II. — Given the plan of the pyramid and the length 
of its axis. 

Draw X X (Fig. 33) parallel to V, the plan of the axis ; 
through V draw V V perpendicular to X X, and through 
draw 0' perpendicular to XX. With 0' as centre and the 
given length of the ax's as radius describe an arc cutting 
V V' in V; then v V will be the height of the pyramid, and 
we now proceed as in Case 1. 

Or; draw Va; perpendicular to V, and with as centre 
and radius equal to the length of the axis describe an arc 
cutting V a; in a; ; Y x will be the height of the pyramid. 



PROBLEM XV. 

To draw tJie pattern of a frustum of an oblique pyramid. 
Case I. — Given the plan of the frustum and its height. 

Let A B C D D' A' B' C (Fig. 85) be the plan of the frustum 
(here of a square pyramid). Produce A A', B B', &c., the 
plans of the edges to meet in a point V ; this point is the 
plan of the apes of the pyramid of which the frustum is a 
part. Join O, the centre of the square which is the plan of 
the large end of the frustum, to V. The line O Y will pass 
through o\ the centre of the plan of the small end ; O 0' will 
be the plan of the axis of the frustum, and V the plan of 
the axis of the pyramid of which the frustum is a portion. 

Draw X X parallel to O V ; through Y draw Y V' perpen- 
dicular to XX, and cutting it in v. Make vx equal to the 
given height of the frustum, and through x draw xx parallel 
to XX; through O draw OQ perpendicular to X X and 



'£[IE TINSmiHS' PATTERN MANUAL.- 



i-iO 



meeting it in Q and through. 0' draw 0' Q,' perpendicular 
to X X and catting xx ia e'. Jx)iu Q e' and produce it to 
intersect v V in V. Next make v a, v b, v c, v d equal to V A, 
V Br V C, V D respectively ; join a. h, c, and d to V by lines 
cutting a; a; in pointg a',b',c', and d'; aa\ hh\ &c., are the 
Lengths of the edges of the frustum. 

Fig. 35. 




To draw the pattern with the seam at A A*. Draw Y A 
(Fig. 36) equal to V a (Fig. 35) ; with V as centre and V b (Fig. 
35) as radius descriLe an arc 6, and with A as centre and A "B 
(Fig. 35) as radius descrihe an arc intersecting arc 6 in B ; 
tvith Vc (Fig, 35) as radius and V as centre describe arc r,. 
and with BO (Fig. 35) as radius and B as centre describe 
an arc intersecting the arc c in C. Kest v/ith V d and V a 
CFig. 35) as radii and V as centre descrilie arcs d and a ; with (J 
as centre and radius CD (Fi^r, 35) describe an arc intersecting 
arc d \nJy ; and with D A (Fig. 35) as rsidius and D as centre 
describe an arc intersecting the arc a in A. Join A^B, C. D, 



150 



TUE TlNSMHiiS' FATTERN MANtJAL, 



and A to V; make A A', E B'', CO', DT>' respectively eqtzrt] 
to a a', h h', cc', d 3; (Fig, 86), aBd join A B, B 0, C D, D A, 
A'B', B'C C'F, &c. ^Thon. A BCD AA; D'C'B A' is tb© 
pattern require. 




A ^ 

(78.) TBe dotted circles (Fig. 35) tbrotigli the angular 
points o£ the plans of the ends show the plans of the ends of 
the frijstuni of the oblique cone which -u'ould envelop the 
frustum of the pyramid. From the simiiarit}'- of the con- 
struction above to that for the pattern of a frustum of an 
oblique cone, it will bo evident that we have treated the 
etlges of the frustum as generating lines (see h, p. 126) of the 
friistum of the oblique cone in which the frustum of the 
pyramid could be inscribed. 

Should it be inconvenient to draw X X in conjunction with 
the plan of the pyramid dr^w X X quit© apart, and from any 



THli' TINSMITHS PATIEKN MANUAL 



151 



point V in it draw v V perpendicular to X X : mate v x equal 
to the height of the frustum and draw xx parallel to X X. 
Make va,vb,vc,vd equal to V A, V B, V C. Y D (Fig. 35 i 
respectively ; and make x a', x h', x c', x d' equal to V A', V B', 
V C, V D' (Fig. 35) respecti vely. Join a a\ h b\ c c', and d d' 
by lines produced to meet u V in V, and proceed ae stated 
above. 

Case IT. — Given the dimensions of the two ends of the 

frustum, the slant of one face and its inclinatioQ (the 

slant of the face of a frustum of a pyramid is a line 

meeting its end lines and perpendicular to them^. 

Draw (Fig. 37) a line E E" equal to the given slant, make 

the angle E" E E' equal to the given inclination, and let fall 

E"E' perpendicular to E E'. Draw A B C D (Fig. 35), the 

plan of the large end of the frustum, and let B C be the plan 

of the bottom edge of the face whose slant is given. Bisect 

B C in E and draw E E' perpendicular to B C and equal to 

Fig. 37. 




E E' (Fig. 87). Through E' draw B' C parallel to B C ; make 
E' C and E' B' each equal to half the length of the top edge 
of the B face, through C and B' draw C D' and B' A' 
parallel to D and B A ; make C D' and B' A' each equal to 
B'C; join D' A', also A A', BB', C C, and D D' ; this will 
complete the plan of the frustum. E' E" (Fig. 37) is the height 
of the frustum. The remainder of the construction is now 
the same as that of Case 1. 



152 



THE TINSMITHS' PATTERN MANUAL. 



For large "work and whore the ends of a frustum are of 
nearly the same size, it would be inconvenient to use long 
radii. For unequal-tapering bodies which are not portions 
of oblique pyramids, as in Problem XVII., the method now 
given, or & modification of it^ must be used. 



PEOBLEM XYI. 

To draw, without ioag radii, the pattern for a frustum of an 
oblique pyramid. The plan of the fnisium and its heujht 
being given. 

Let A B C D D' A' B' C (Fig. 38) be the plan of the frustum. 
From any point S in B C draw E E' perpendicular to B C and 
B' C, the plans of the bottom and top edges of the face B 
B' C of the frustum. Draw E'.E" perpendicular to E E' aad 

Fig. 33. 




eoual to tiie height (which either is given or can bo found 
as in Case IL of last problem), and join E E", then E E" is the 
true length of a elant of the face B C B' C, of the frustum. 
Join D €!',, and find its true length (D C") by drawing C G" 



THE TINSMITHS' PATTERN MANUAL. 



m 



perpendicTjlar to D C and equal to the height of frustum and 
joining D C". Kest join D' A and B' A ; through D' and B' 
draw lines D' A", B' B" perpendicular to D' A and B' A 
respectively, and naako D' A" and B' B" each equal to the 
given height of the frustum ; join A A" and A B", then A A" 
and A B" are the true lengths of D' A and B' A respectively. 
To draw the pattern of tlie face B"CB'C',. draw E E' 
(Fig. 89) equal to E E" (Fig. 38), and through E and E' draw 
B C and B' O' perpendicular to E E'. Make E C, E B, E' C, 
and E' B' equal to E C, E B, E' C, and F' B' (Fig. 38) respec- 
tively; join C and BB'; this completes the pattern of tbe 
face. The patterns of the other faces ai-e found in the 
following manner:— "With C (Fig, 39) and as centres and 




D C" and C D (Fig, 38) as radii renpectivelyj deecribe arcs 
intersecting in D; join CD, draw CD' parallel to CD and 
equal to CD' (Fig. 38); and join D D', With D' and D 
(Fig. 39) as centres and A A" and DA (Fig, 38) as radii 
respectively describe arcs intersecting in A ; join D A, draw 
B' A' parallel to D A and equal to D' A' (Fig. 38), and join A 



154 



THE TINSMITHS' PATTERN MANUAL. 



to A'. Next, •witli B' and B as centres and A B" and B A 
(Fig. 38) respectively as radii/describe arcs intersecting in A ; 
^'oin B A- and draw B' A' parallel to B A and equal to B' A' 
(Fig. 88). Join A A', and this will complete the pattern 
required. 



PEOBLEM XVII. 

To draio (he pattern for a "hood. 

The plan of the hood is necessarily given, or else 
dimensions from which to drav/ it. Also the height of the 
hood, or the slant of one of its faces. The hood is here 
supposed to he a hody of unequal taper with top and base 
parallel, but not a frustum of an oblique pyramid. 

Let A B C D A' B' C D' (Fig. 40) be the given plan of the 




hood (a hood of three faces), A D being the * wall line,' A B 
and D C perpendicular to A D and B C parallel to it, also let 
the length of F C", a slant of face BB'O'C, be given. 
Draw C F perpendicular to B C and through C draw C C" 
perpendicular to C'F, and with F as centre and radius 



THE TINSMITHS' PATTERN MANUAL. 



155 



equal to tlie given lengtli, describe an arc cutting C C" 
in C". Join F C" ; then C G" is the height of the hood, 
which "we need. If the height of the hood is given instead 
of the length F C", make C C" equal to the height and 
join F C", which will be the true length of F C Next, 
through C draw C E perpendicular to CD; draw'C'C" 
pei^ndicular to C E, make C C" equal to the height and 
join EC". Now produce C B' to meet AB in G; draw 
B' B" perpendicular to B' G and equal to the height, and 
join G B". 

To draw Ihe pattern of the hood. Draw F C (Fig. 41) 
equal to F C" (Fig. 40) ; through F and C draw B C and 
B' C, each perpendicular to F C ; make F B equal to F B 
(Fig. 40) ; make F G equal to F G (Fig, 40), and C'B' equal 
to C B' (Fig. 40). Join B B' and C C, then B B' C' C wHl be 
the pattern of the face of which B B' C' G (Fig. 40) is the 
plan. To draw the pattern of the face C D' D C (Fig, 40). 
With C' and C (Fig. 41) as centres and EC" and CE 

Fig, 41. 




(Fig, 40) as radii respectively, describe arcs intersecting in E. 
Join G E and produce it, making C D equal to G D (Fig, 40), 
and through C draw C D' parallel to C D and equal to C D' 
(Fig. 40). , Join D D', then C C D D' is the pattern of the face 
of which a ODD' (Fig. 40) is the plan. With B' and B as 



156 THE TINSMITHS' PATTERN MANUAL. 

centres and radii respectively eqnal to B" G and B G (Fig. 40), 
describe arcs intersecting in G. Join B G and produce it, 
making B A equal to B A (Fig. 40), and throngli B' draw 
B' A' parallel to B A and equal to B' A' (Fig, 40). Join A A' ; 
and the pattern for tlie hood is complete. 



THE TINSMITHS'. PATTERN xMANUAL.' 157 



CHAPTER IV. 

Patteens for Unequal-tafeeing Articles of Flat and 
Curved Surface combined. 

Class II, [Subdivision c.) 

From what lias been stated about the plans of unequal- 
tapering bodies and from g, p. 129, it will bo evident that the 
curved surfaces of the articles now to bo dealt with are 
portions of f i usta of oblique cones. 

(79.) The advantages referred to in § 61 of Ijooking upon 
the oblique cylinder as frustum of an oblique con© will 
be evident in this chapter. For there is to each of the 
problems a Case where the plan arcs of the curved portions 
of the body treated of have equal radii. To deal with these 
as problems excej^tional to a general principle would be 
most inconvenient. . j*3 extreme cases, however, of the one 
principle that the curved portions of the bodies before us are 
portions of frusta of oblique cones, their solution presents no 
difficulty. It will be sufficient to take one such Case in 
connection with only one of the bodies. This we shall do in 
'^Ge IV. of the aext problem. 



PROBLEM XVIII. 

To draw (fie pattern for an unequal-tapering body with top and 
base paralMl and Jiaving flat sides and semicircular ends (an 
' equal-end ' bath, for instance'), the dimensions of top and 
bottom of the body and its height being given. 

Five cases will bo treated of; four in this problem and 
One ia the problem following. 



15S 



THE TINSMITHS' PATTERN MANUAL. 



Case I.™Pattenis wlien the body is to "ba mado i:p oi fciir 

pieces. 
Draw (Figo 42) the plan of the body (see Problem VI,, 
p. 130), preserving of its construction the centres 0, 0' and the 
points A, A' in which the plan lines of the sides and curves 

Fig. 42. 




A G 



I' A 



of the ends meet each Obhsr, Join A A'j as shown (four 
places) in the fig. The ends A D A A' D' A' and A E A A' E' A' 
of the body (see g, p. 129) are portions of frusta of oblique 
cones. Let us suppose that the seams are to be at the four A 
corners where they are usually placed, and to correspond 
with the four lines A A'. Then we shall require one pattern 
for the flat sides, and another for the semicircular ends. 

To draw the end pattern, 

(80.) Draw A 6 D D' A' (Fig. 43) the A 5 D D' A' portion of 
Fig. 42 separately, thus. Draw any line S X and with any 
point (to correspond with 0, Fig, 42) in it as centre and 
O D. (Fig. 42) as radius describe an arc (hero a quadrant) Dh A. 
equal to the arc D & A (Fig. 42). Make D O' eqiial to D 0' 
(Fig. 42), and with 0' as centre and 0' D' (Fig. 42) as radius 
describe an arc (here a quadrant) D' A' equal to the arc D' A' 
(Fig. 44). Joining A A' completes the portion of Fig. 42 
required. Now divide D A into any number of equal parts, 
here three, in the points & and c. From D' draw D' D" per- 
pendicular to X X and equal to tho given height. Then 
P, D" are, in eieYation, the corresponding points of which 



THE TINSMITHS' PATTERN MANUAL. 



159 



D, D' are the plans. Being corresponding points, they are 
in one and the same generating line (§ 68). Join D D" and 
produce it indefinitely, then somewhere in that line will lie 
the elevation of the apex of the cone of a portion of which 



Fig. 43. 




A & D D' A' is the plan. Now from 0' draw 0' 0" perpen- 
dicular to X X and equal to the given height.. Then 0, 0" 
are, in elevation, the centres of the ends of the frustum in 
the same plane that D, D" are represented in, that is in the 
plane of the paper ; 0, 0' being the plans of these centres. 
Join 0" and produce it indefinitely ; then in this line lies 
the elevation of the asis of the cone of a portion of which 
A fe D D' A' is the plan, and necessarily therefore the elevation 
of the apex. That is, the intersection point P of these two 
lines is the elevation of the apex. Next, from P let fall P P' 
perpendicular to X X ; then P' will be the plan of the apex. 
Join D" 0". With P' as. centre and P'c, P' b, and P' A sue- 



160 



THE TINSMITHS' PATTERN MANUAL. 



cessively as radii, describe arcs cutting XX in C, B, and A". 
Join these points to P by lines cutting 0" D" in C, B', and A'. 
Next draw a line P D (Fig. 44) equal to P D (Fig. 43), and 
with P as centre and P C, P B, and P A" (Fig. 43} successively 
as radii describe arcs cc.hh, and a a. With D as centre and 
ladius equal Do (Fig. 43) describe arcs 'cutting arc c c in C 
,and C right and left of P D. With same radius and these 
jii-mts C and C successively as centres describe arcs cutting 
arc bh in B and B ri-ht and left of P D. With B and B 




successively as centres and same radius describe arcs cutting 
arc a a in A and A right and left of P D. Join the points 
C, B, and A right and left of P D to P. With P as centre 
and P U" (Fig. 43) as radius describe an arc cutting P D iu 
D". With same centre, and P C (Fig. 43) as radius, describe 
arc c'c' cutting lines P C right and left of PD in C and C. 
With same centre, and P B' (Fig. 43) as radius, describe arc 



THE TINSMITHS' PATTERN MANUAL.^ 16I 

h' V cutting lines P B right and left of P D in B' and B'. 
Similarly find points A' and A'. Through the successive 
points A, B, C, D, C, B, A, draw an unbroken curved line. 
Also through the successive points A', B', C, D', C, B', A, 
draw an unbroken curved line. Then AD A A' D' A' will 
be the required pattern for ends of the.body. 

To draw the pattern for the sides. 

" Through A' (Fig. 42) draw A' F perpendicular to A' A' 

make F G equal to the given height and join A' G. Then 

A' G is the slant of the Jbody at the side. Next draw 

(Fig. 45) A A equal to A A (Fig. 42), and mate A F equal 

Fig. 45. 




to A F (Fig. 42) ; through F draw F A' perpendicular to A A 
and equal to.A'G (Fig. 42), and through A' draw A' A' 
parallel to A A. Make A' A' equal to A' A' (Fig. 42). Join 
A. A', A A', then A A' A' A is the pattern for the sides. 

Case II. — Pattern when the body is to be made up of two 

pieces. 

We will take it that the seams are to be at D D' and E E^ 
_(Fig. 42). It is evident that we want but one pattern, which 
shall include a side of the body and two half-ends. 

First draw as just explained A' A F A A' (Fig. 46) a f-ide- 
pattern of the body. Produce one of the lines A A' of this 
pattern, and make A P' equal to A"P (Fig. 43). With P' as 
centre and P B, P C, and P D (Fig. 43) successively as radii 
describe arcs h, c, and d, and with A as centre and A h 
(Fig. 43) as radius describe an arc cutting arc 6 in B. With 
same radius and B as centre describe an arc cutting arc c 
in C ; similarly with C as centre and same radius find D.( 



162^ 



THE TINSMITHS' PATTERN MANDAL. 



J'oin B F, C P', D P'. Now with P B' (Fig. 43) as radius and 
P' as centre describe an arc. h' cutting PB in B', and with 
P C, P D" (Fig. 43) successively as radii describe arcs c' and 
d cutting P' and P' D in C and D'. Through the points 



Fi(J. 46. 




A, B, C, and D draw an unbroken curved line. Also through 
the points A', B', C, and D' draw an unbroken curved line. 
Then A D J)' A' will bo a half-end pattern attached to the 
right of th^ side pattern. Draw the otber half-end pattern 
A E E' A' in the same manner ; then E F D D' A' A' E' will be 
the^omplete pattern required. 

^ASB ni. — Pattern when the body is to bo made up of one 
^ piece. 

>Tn this case we will put the seam to correspond with D D' 
(Fig. 42).' 

\: First draw a E A A' E' A' (Fig. 47) an end pattern of the 
body in the same manner that A D A A' D' A' (Fig. 44) was 
arawn. ^ "With A' and A (right of E E') as centres and A' G 
aniAF (the small length AF) (Fig. 43) respectively aa 



THE TI^ii^MITHS' PATTERN MANUAL. 



163 



radii desirilDO arcs intersecting in F; join AF and pi'odiice 
it, making A A equal to A A (Fig. 42). Through A 
{extremity of F A') draw A' A' parallel to A A and equal to 




Ui 



THE TINSMITHS' PATTERN MANl'AL, 



A' A' (Fig. 42). Join A A', the extremities of the lines just 
drawn, and produce it indefinitely ; and make A P' equal 
A" P (Fig. 43). With P' as centre and P B, P C, and P D 
(Fig. 43) successively as radii describe arcs b, c, and d, and 
with A (of A P') as centre and A b (Fig, 43) as radius describe 
an arc cutting arc b in B. With same radius and B as centre 
describe an arc cutticg arc c in C, and similarly with C as 
centre and same radius find D. Join B P', C F, and D P'. 
Nsw with P' as centre and radii successively equal to P B', 
PC, and^PD" (Fig. 43) describe arcs h', c', and d' cutting 
P'B, FC, and FD in B', C, and D'. Through the points 
A, B, C, and D draw an unbroken curved line. Also through 
the points A', B', C, and D' draw an unbroken curved line. 
We have now in D F A A' A' D' attached to the right of the 
end pattern we started with, a side pattern and a half-end 
pattern. By a repetition of the foregoing construction we can 
attach A A D D' A' A' to the left of the end pattern we 
started with. The figure D E D D' E' D' will be the completo 
pattern required. 

Case IV.— Where the plan arcs D A, D' A' (Fig. 42) have 
equal radii. 
This is the extreme case above (§ 79, p. 157) referred to, 
where the cone becomes cylindrical. Problem V., p. 121, may 

Fig. 48. 

A* ,. S' CD' 




advantageously be compared with th& work now given. The 
arcs (Fig. 48) D A and D' A' (here quadrants) being equal, 



THE TINSMITHS' PATTERN MANUAL. 



1G5 



tlieir radii D, 0' D' are ako equal. Through D' and 0' draw 
D' D", and 0' A" perpendicular to X X and each equal to the 
given height of the body. Join D D", A" D" ; divide the arc 
D A into any number of equal parts, here three, in the 
points h and c; and through c, h, and A draw cC, t B, and 
A each perpendicular to X X and cutting it in C, B, and O 
respectively. The arc D A being here a quadrant the point 
where the line from A perpendicular to X X cuts X X is 
necessarily O, the centre whence the arc is drawn. Through 
C, B, and draw C C", B B", and O A" parallel to D D" and 
cutting A" D" in points C", B", and A". Also through D" 
draw a line D"A' perpendicular to DC and cutting the 
lines just drawn in C, B' and A'. Make c 2 equal to C c ; 
B' 1 equal to B b, and A' equal to A. From D" through 
2, 1, to O draw an unbroken curved line. 

To draw the pattern. 

Draw D D" (Fig. 49) equal to D D" (Fig. 48) and through 
D" draw au indefinite line A' A' perpendicular to DD". 

Fig, 49. 




Mark off on A' A', right and left of D", D" C, C B', and B A' 
respectively equal to the distances between D" and 2, 2 and 
1, and 1 and (Fig. 48) ; and through C, B', and A', right 
and left of D" draw indeSnite lines each parallel to D D". 
Make CC right and left of D D" equal to C C (Fig. 48) ; and 
make B' B right and left of D D" equal to B'B (Fig. 48). 
Also make A' right and left of D D" equal to A' (Fig. 48). 



166 



THE TINSMITHS' PATTERN MANUAL. 



Next make C'C" right and left of DD" equal to C'C' 
(Fig, -48). Similarly find points "B", A" right and left of 
D D" by making B' B", A' A" respectively equal to B' B", and 
A' A" (Fig. 48). Through the points 0, B, C, D, C, B, O, 
draw an unbroken curved line. Also through the points 
A", B", C", T>", C'\ B", A", draw an unbroken curved Hne. 
Then D A" D" A" will be the pattern required* 



PEOBLEM XIX. 

To draw, without long radii, the paUem for ai unequal' 
tapering hodp with top and base parallel and having fiat sides 
and equal semicircular ends {an ' equal-end ' bath, for instance). 
The dimensions of the top and hottom of the body and its 
height being given. 

This i problem is a fifth ease of the preceding, and is 
exceedingly useful where the work is so large that it is 
inconvenient to draw the whole of the plan, and to use long- 
radii. 

To draw the pattern. 

Fig. 50. 




First draw half the plan (Fig. 60). It is evident that the 
df awing of the side pattern presents no difficulty, as long 



THE TINSMITHS' PATTEHN MANUAL. 167 

radii are not involved. It can be drawn as in Case I. of 
preceding problem. Divide the quadrants D A; D' A', each 
into the same number of equal pirts, here three, in the points 
c, h, c', b' ; join c c', h b'. Through D' draw D' E perpendicular 
to D' D and equal to the given height of the body. From 
D' along D' D mark off D' A, D' B, and D' C respectively 
equal to A' A, b' b, and c' c ; and join points D, C, B, and A to 
E, then E A, E B, E 0, and E D, will be the true leng-hs of 
A' A, b' b, c' c, and D' D respectively. Next join c D', and 
draw D' d" perpendicular to D' c and equal to the given 
height. Join c d", then cd" may be taken as the true length 
of D'c. Similarly join be' and A &', and through c' and V 
draw c' c" and b' b" perpendicular to c' b and 6' A respectively, 
and each equal to the given height. Join b c"and A b", then 
b c" and A b" may be taken as the true lengths of h c' and A 6' 
respectively. 

Now draw (Fig. 51) a line DD' equal to D E (Fig. 50) and 
with D' and D as centres and rndii respectively equal to d" c 




and Dc (Fig.' 50) describe arcs right and left of DD', inter- 
secting in c and c. With c, right of D, and D' as centres, and 
radii respectively equal to C E and D' c' (Fig. 50) describe 
arcs intersecting in c\ right of D'. With c, left of P, and D' 
as C(!ntres, and radii as before, describe arcs intersecting in c', 
left of D'. With successively c' and c right and left of D D', 



i6§ TllB TINSMITHS' PATTERN MANUAL. 

as centres and radii respectively eqttal to c" 6 and c h (Fig. 50) 
describe arcs intersecting in h and 6 right and left of D D'. 
With successively 6 and c' right and left of D D' as centres and 
radii respectively equal to B E and c' h' (Fig. 50) describe area 
intersecting in h', and 6' right and left of DD', and with suc- 
cessively 6' and 6 right and left of D D' as centres and radii 
respectively equal to A b" and 6 A (Fig. 50) describe arcs in- 
tersecting in A and A right and left of D D'. Similarly with 
A E and b' A' (Fig. 50) as radii and centres respectively A 
and b' describe intersecting arcs to find points A and A', right 
and left of D' D. Through the points A, b, c, D, c, b, A draw 
an unbroken curved line. Also through the points A', b', c 
jy, c\ b', A', draw an unbroken curved line. Join A A', A A', 
right and left of D D', then A D A A' D' A' will be the pattern 
required. 

The lines c c\ c D', 6 6', &c., are not needed in the working ; 
they are drawn here to aid the student by showing him how 
the pattern corresponds with the plan, line for line of same 
lettering (see also § 66, p. 121). 



PROBLEM XX. 

To draw the pattern for an oval unequal-tapering body with top 
and base parallel {an oval bath, for instance). The height 
and dimensions of the top and bottom of the body being given. 

Four cases will be treated of, -three m this problem, and 
one in the problem following (see also § 79, p. 157). 

Draw (Fig. 62) the plan of the body (see Problem VII., 
p. 131), preserving of its construction, the centres 0, O', P, V 
and the several points d and d' in which the side and end 
curves medt each other. Join dd', as shown (four places) in 
the fig. From the plan we know (see g, p. 129) that dGd 
d! Q' d/f d'Bdd'W d', the ends of the body are like portions 
qf the frustum of an oblig^ue cone; we also know that 



!fHE TINSMITHS' PATTERN MANUAL; 



169 



dAdd' A'd', dEd d''E' d', tLe Bides of the "body are like por- 
tions o£ the frustum of an oblique cone. 
In Plate I. (p. 181), is a representation of the oval unequal- 









Fia 


52. 
JS 








/ 




/^ 


" ' 


i" 


~>^ 




\ 


^ 










I 


' 


u 

is 


\ 


r 

V 






o' 


'(? 


; 


/ 


N 


V 






"6^ 




>/ 


i 



tapering body for which patterns are required, also of two 
oblique cones {x and Z). The oblique cones show (except as 
to dimensions) to what portions of their surfaces the several 
portions of the surface of the oval body correspond. Thus 
the sides, A', of the body correspond to the portion A of cone 
a;, and the ends, B', B', correspond to the B portion cone Z. 
The correspondence will be more fully recognised as we 
proceed with the problem. The difference of obliquity 
between B' and B is seeming only, not real ; and arises simply 
from Z being turned round so that the whole of the dV> d(X 
B' ^ (Fig. 52) of the cone shall be seen. If the representation 
of Z showed its full obliquity, then the line on it from base 
to apex would be the right-hand side line of the cone, and 
only half of the B portion could be seen. 




x: ' 



THE TINSMITHS' PATTERN MANUAL. 171 ■ 

Case L — Patterns when the body is to be made up of 
four pieces. 

It is clear that we require two patterns ; one for the two 
(ends, and one for the two sides ; also that the seams should 
correspond with the four lines d d', where the portions of the 
reepective frusta meet each other. 

To draw the pattern for the ends. 

Draw separately (Fig. 53) the G'Gfdd' port 'on of Fig. 52, 
thus. Draw any line X X, Fig. 53, and with any point O 
(corresponding to O, Fig. 52 j in it as centre and O G (Fig. 52) 
as radius, describe an arc G d equal to G d of Fig. 52. Make 
G 0' equal to GO' (Fig. 52), and with 0' as centre and O' G' 
(Fig. 52) as radius describe an arc G' d' equal toG'cZ' of 
Fig. 52. Joining d d' completes the portion of Fig. .52 required. 
Now divide the arc G d into any number of equal parts, here 
three, in the points / and e. At G' and 0' draw G' G", O' O" 
perpendicular to X X, and each equal to the given height of 
the body. Join O O", G G" ; produce them to their intersection 
in S (§ 80, p. 158) ; and from S let fall S S' perpendicular to 
XX. Join 0"G". With S.' as centre and S'/, S'e, and S'd 
Buccessively as radii, describe ai;cs cutting X X in F, E, and D. 
Join these points to S bylines cutting 0" G""in F', E', and D'. 
- Next draw S G (Fig. 54) equal to S G (Fig, 53), and with 
S as centre, and SF, SE,and SD (Fig. 53) successively as 
r::dii describe arcs //, e e, ami d d. With G as centre and 
xadius equal to G/ (Fig. 53) describe arcs cutting arc // 
right and left of S G in F and F. With each of these point.s, 
F and F as centre and same radius describe arcs cutting arc 
e e right and left of S G in E and E. With same radius and 
each of the last-obtained_j)oint8 as centre describe arcs cutting 
dd jight and left of SG in D and D. Join all the points 
right and left of SG to S. With S as centre and SG" 
(Fig. 53) as radius, describe an arc cutting S G in G'. With 
Fame centre and S F' (Fig. 5^ as radius, describe an arc ff 
CTitting the lines S P right^nd left of S G in F' and F'. 
With same centre and S E' (^ig. 53) as radius, describe an 



172 



Fig. 54. 




THE TINSMITHS' PATTERN MANUAL. 



173 



arc e' e' cutting the lines S E right and left of S G in E' and E'. 
Similarly by arc d' d' obtain points D' and D'. Through the 
points D, E, F, G, F, E, D, draw an unbroken curved line. 
Also through points D', E', F', G', F', E', D', draw an unbroken 
curved line. Then D G D D' G' D' will be the required pattern 
for the ends of the body. It is in fact the development of 
the B portion of cone Z of Plate I. 

To draw the pattern for the sidss. 

Draw separately (Fig. 55) the ' Ahdd' portion of Fig, 52, 
thus. Draw any line X X, Fig. 55, and v/ith any point P 

Fig. 53. 




(corresponding to P, Fig. 52) in it as centre and P A (Fig. 52) 
^ radius, dcGcribe an arc A d equal to A <? of Fig. 52, Make 



:i74 



TUE TINSMITHS' PATTERN MANUAL. 



.AF equal to A-F (Fig. 62) and witli F as centre and F A' 
(Fig. 52) as radius describe an are A! d' equal to K' d' of 
Fig. 52. Joining d d' completes the portion of Fig. 52 required. 
Now divide the arc A d into any number of equal parts, here 
three, in the jpoints b and c. At A' and P' draw A' A", P' P" 
perpendicular to X X, and each equal to the given height 
of the body. J^in P P", A A", produce them to their inter- 
section in Q (§ 80, p. 158) ; and from Q let fall Q Q' perpen- 
dicular to XX. Draw a line through P"A". With Q' as 
centre and Q' 6, Q' c, and Q' d succ ssively as radii describe 
arcs cutting X X in B, C, and D. Join these points to Q by 
Hues cutting that through P" A", in B', C, and D'. • 

Next draw Q A (Fig. 56) equal to Q A (Fig. 55), and with 
Q as centre and Q B, Q C. and Q D (Fig. 55). successively as 

Fig. 5G. 




radii describe arcs hh, cc, and dd. With A as centre and 
radius equal to A 6 (Fig. 55) describe arcs cutting arc h h 
right and left of Q A in B and B. With each of these points 



THE TINSMITHS' PATTERN MANUAL. 175 

B and B as centre and same radius describe arcs cutting arc 
c c right and left of Q A in C and 0. With same radius and 
each of the last-named points a;s centre describe arcs cutting 
arc d d right and left of Q A in D and D. Join all the points 
right and left of Q A to Q. With Q as centre and Q A" 
(Fig. 55) as radius, describe an arc cutting Q A in A'. With 
same centre, and Q B' (Fig. 55) as radius, describe an arc 6' & 
cutting the lines Q B right and left of Q A in B' and B'. 
With same centre and Q C (Fig. 55) as radius, describe arc 
c' c' cutting the lines Q C right and left of Q A in C and C. 
SimiLirly by arc d' d' obtain points D' and D'. Tiirough the 
points D, C, B, A, B, C, D, d,raw an unbroken curved line. 
Also through the points- D', C, B', A', B', C, D', draw. an un- 
broken curved line. Then D A D D' A' D' will be the required 
pattern for the sides of the body, and is in- fact the develop- 
ment of the A portion of cone x of Plate I; 

Case II. — Pattern when the body is to be made up of two 

pieces. 

In this case the seams are usually made to correspond with 
B B' and G G' (Fig. 52). It is evident that only one pattern 
is now required, made up of a pattern for the side A' of the 
body (Plate I.) with right and left a half-end (B',JB', Plato I.) 
pattern attached. 

Draw (Fig. 57) a side pattern D A D D' A' D' as described 
in C..S6 I. Produce D Q and make D S equal to D S (Fig. 53), 
With S as centre and S E, S F, and S G.(Fig. 53) successively 
as radii describe arcs e, f, and g, and with D as centre and 
de (Fig. 53) as radius describe an arc cutting arc e in E. 
With (-amo radius and E as centre describe an arc cutting 
arc / in F, and similarly with F as centre and same radius 
find G Join E S, F S, , G S. Now with S E' (Fig. 53) as 
radius and S as centre describe an arc e' cutting S E in E', 
and with S F' and S G" (Fig. 53) successively as radii describe 
arcs /' and g' cutting S F and S G in F' and G'. From points 
D to G draw an unbroken curved line. Also from points D' 
to G' draw an unbroken curved line. Draw the other half-end 



176 



Via. 57. 





r — r 


1, 


\1 




/ 



THE TINSMITHS' PATTERN MANUAL. 177 

pattern D B B' D' in tlie same manner ; then G A B B' A' G ' 
will be the pattern required. 

Case III. — Pattern when the body is to be made up of one 

piece. 

"We will put the seams at the middle of one end of the 
body, say, to correspond with B B' (Fig. 52). We now need 
an end pattern (the end d G d d' G' d' in plan), with side 
pattern attached right and left (dEdd'l&'d', dAdd'A'd' 
in plan), and attached to each of these a half-end pattern 
((* B B' d', dBB' d' in plan). For want of space we do not 
give the pattern, but it is evident from what has just been 
stated, that the pattern will be double that shown in Fig. 57. 
It will be a useful exercise and should present no difficulties 
to the student, ■lo himself draw the complete pattern, first 
drawing an end pattern (see Case I. and Fig. 54) and attaching 
right and left, a side pattern and a half-end pattern, 

PBOBLEM XXI. 

To draw, without long radii, the pattern for an oval unequal- 
tapering hody, with top and base parallel (an oval bath, for 
instance). The height and the dimensions of the top and 
bottom of the body being given. 

This problem is a fourth case of the preceding, and will be 
found very useful for both the end and side patterns, the 
radii of which are often of a most inconvenient length. 

To draw the end pattern. 

First draw (Fig. 68) the plan of the end of the body; that 
is the dG dd' G d' portion of Fig. 52. Divide the arcs G d, 
G' d' each into the same number of equal parts, here three, in 
the points /, e, /', e' ; join //', ee'. Through G draw GH 
perpendicular to G G' and equal to the given height of the 
body. From G along GG' mark off G-F, GE, and GD re- 
spectively equal to //', e e', and d d' ; join G', F, E and D to 11 ; 
then G' H, F H, E H, and D H will be the true lengths of 



178 



THE TINSMITHS' PATTERN MA-NUAL. 



GG', //', ee', and dd' respectively. Next join /G'.; draw 
G'/ perpendicular to/G' and equal to the given height and 
join//; then fg" may be taken as the true length of G'/. 
Similarly join ef\ d q\ and through / atid e' draw /'/" and 



Fig. 58. 




e' e" perpendicular to f e and e^ d respectively, and each equal 
tu the given height, and join e/"and d e" ; then ef" and de" 
may l)e taken as the true lengths of ef and d e' respectively. 
Next draw (Fig. 59) a line G G' equal to G' H (Fig. 58) 
and with G' and G as centres and radii respectively equal to 
fg" and G/(Fig. 58) describe arcs right and left of G G', 
intersecting in / and /. With /, right of G, and G' as centres, 
imd radii respectively equal to F H and G'/' (Fig. 58) describe 
j.rcs intersecting in /', right of G'. With /, left of G, and G' 
us centres, and radii respectively as before, describe arcs 
intersecting in /, left of G'. With successively /' and /, right 
:nid left of G G', as centres and radii respectively equal to 
cf" and fe (Fig. 58) describe arcs intersecting in e and e. 
With successively e and /' right and left of G' G as centros 
'and radii respectively equjil to E 11 and /' e' (Fig. 58) deBcrib<- 



THE TINSMITHS- PATTERN MANUAL. 



179 



arcs intersecting in e' and e' ; and with successively e' and e 
right and left of G G' as centres, and radii respectively equal 
to d e" and e d describe arcs intersecting in d and d. Also 
with successively d and e' as centres and D H and e' d' re- 
spectively as radii describe arcs intersecting in d' and d'. 

Fig. 59„ 




END PATTERN. 



Throtigh d, e, /, G/, e, d, draw an unbroken curved line. 
Also through d\ e\ /', G', /', e', d\ draw an unbi oken curved 
line. Join d d', right and left of G G', then dGdd' G'd' will 
be the end pattern required. 

The lines //', e e', G'f, f e, &c., are not needed in the work- 
ing, they are drawn for the reason stated in § 65, p. 121. 

To draw the side pattern. 

First draw (Fig. 60) the plan of the side of the body ; that 
is the d Ad d' A' d' portion of Fig. 52. Divide the arcs A d, 
A' d' each into the same number of equal parts, here three, 
in the points h, c, b\ c' ; join & V, c c'. Through A draw A E 
perpendicular to A A' and equal to the given height of the 
body. /Fiom A along A A' mark off A B, AC, and AD re- 
spectively equal to h h', c c', and d d'. Join A', B, C and D 
to E ; then A' E, B E, C E, and D B will be the true lengths 
of A A', b b', c c', and d d' respectively. Next join h' A and 
draw b' b" perpendicular to b' A and equal to the given heit-lit. 



180 THE TINSMITHS' PATTERN MANUAL. 

Join A h" ; tKen A h" may he talcen as tlie true length A h'. 
Similarly join c' b and d' c ; througli c' and d' dra-w c' c", and 
d' d'' perpendicular to c' 6 and d' c respectively, and each 
equal to the given height ; join h c", c d", then h c" and c d" 
may be taken as the true lengths of 6 c' and c fTjrespectively. 

Fig. 60. 




Kext draw (Fig. 61) a line A A' equal to A'E (Fig. 60) 
and with A and A' as centres and radii respectively equal to 
Ah" and A' b' (Fig. GO) describe arcs right and left of A A', 
intersecting in b' and b'. With h', right of A', and A as 
centres, and radii respectively equal to B E and A b (Fig. 60) 
describe arcs intersecting in &, right of A, With b', left of 
A', and A as centre.«, and radii respectively as before, describe 
arcs intersecting in h left of A. With successively b and b' 
right and left of A A' as centres, and radii respectively equal 
to b c" and b' c' (Fig. 60) describe arcs intersecting in c' and c'. 
With successively c' and b right and left of A A' as centres, and 
radii respectively equal to C E and h c (Fig. 60) describe arcs 
intersecting in c and c j and with successively c and c' right 



THfi TINSMITHS' PATTEKN MANUAL, 



18^ 



and left of A A' as centres, and radii respectively eqnal to c d" 
and c' d' (Fig. 60) describe arcs intersecting in d' and d'. 
Also with successively d' and c right and left of A A' as 
centres, and D E and c d respectively as radii describe arcs 



Fig. 61. 




1' 6' ^' h c^d' 

SIDE PATTEKN. 

intersectirsg in d and d. Through d, c, b, A, b, c, d, draw an 
unbroken curved line. Also through d', c\ b', A ', b', c', d\ 
draw an unbroken curved line. Join d d\ right and left of 
A A' ; then d Kdd' A! d' will be the side pattern required. 

The remark about lines //', ee\ &c., in (end pattern) 
Fig. 59, applies to lines e 'c\ b h', b c\ &c., in the present pattern. 



FEOBLEM XXn. 

To draw ilie paUern for a tapering body with top and base 
parallel, arid liamng circular top and oblong bottom with 
semicircular ends (tea-bottle top, for instance), the dimensions 
of the top and bottom of the body and its height being given. 

Four cases will be treated of; three in this problem and 
one in the problem following (see also § 79, p. 157). 

Case I. — Pattern when the body is to be made up of four 
pieces. 

Draw (Fig. 62) the plan of the body (see Problem VIII., 
p; 132) preserving of its construction the centres 0, 0'; the 
points & in plan of bottom where the extremities of the pla« 



184 



THE TINSMITHS' PATTERN MANUAL. 



lines of the sides meet the extremities of the plan semicircles of 
the ends ; and the points 6' in plan of top where the sides and 
ends meet in plan. Join V h at the four corners. The ends 
bBhh'B' V, and feUb 5' E' fe' of the body (see g, p. 129), are 
portions of frusta of oblique cones. Making the body in four 
pieces it will be best that the seams shall correspond with 
the lines A II, B B', D h', and E E', then one pattern only, 
consisting of a half-end with a half-side pattern attached, - 
will be req^uired. 




To draw the pattern. 

Draw separately E' E dh V (Fig. 63), the E' E tZ 6 ^'portion 
of Fig. 62, thus. Draw any line X X and with any point O 
(to correspond with O, Pig. 62) in it as centre and O E (Fig. 62) 
as radius describe an arc (here a quadrant) 'Eidh, equal 
to E J" 5 of Fig. 62. Make E O' equal to E O' (Fig. 62), and 
with 0' as centre and O'E' (Fig. 62) as radius describe an 
arc (here a quadrant) E' l! equal to E' V of Fig. 62. Joining 
h h' completes the portion of Fig. 62 required. Now divide 
E 5 into any number of equal parts, here three, in the points 
d and c.' From E' and O' draw E' E", 0' 0" perpendicular to 
X X and each equal to the given height of the body. Join 
E E", O" ; produce them to intersect in P (§ 8(f, p. 158) ; from 
P let fall P P' perpeudicTilar to X X, and join E" 01'. With P' 



THE TINSMITHS' PATTERN MAXrAl-. 



18c 



as centre and P' d, P' c, and P' h successively as radii describe 
arcs cutting X X in D, C, and B. Join these, points to P by 
lines cutting 0" E" in D', C, and B' 



Fig. 63. 




Next draw (Fig. 64) a line P E eqnal to P E (Fig. 63), and 
with P as centre, and P Dt P C, and P B (Fig. 63) successively 
as radii describe arcs d, c, and 6. With E as centre and 
radius equal to 'Ed (Fig. 63) describe an arc cutting arc d in 
D. With same radius and D as centre describe an arc cutting 
arc c in Cj and with G as centre and same radius describe an 
arc cutting arc 6 in B. Join the points D, C and B to P. 
With P as centre and PE" (Fig. 63) as radius describe an 
arc e' cutting P E in E'. With same centre and P D'.(Fig. 68) 
as radius describe arc d' cutting P D in D'. Similarly 



i86 



THE TINSMITHS' patterin: MANUAL; 



wit] I same centre and P C and P B' (Fig. 63) successively as 
radii find points C and B'. Through h' (Fig. 62) draw h' H 
perpendicular to h' D and equal to the given height, and 
join D H, then D H will be the true length of the line of 
which h' D is the plan, that is, will be the length of a slant 
,of the body at the middle of the side, where one of the seams 

1 Fig. 64. 




will come. With B' and B (Fig, 64) as centres and radii 
respectively equal to D H and 6 A (Fig. 62) describe arcs 
intersecting in A. Through the points E, D, C, and B draw 
an unbroken curved line. Also through the points E', D', C, 
and B' draw an unbroken curved line. Join B A, A B' ; then 
E C A B' E' will b© the pattern re(juired. 



THE TINSMITHS' PATTERN MANUAL. 



187 



Case II, — Pattern when the body is to be made ud of two 

pieces. 

We will suppbse the seams are to correspond with the lines 
B B' and E E'. It is evident that here we need but one 
pattern only, which will combine a side of the body and two 
half-ends, in fact will be double that of Fig. 64. 

First draw B B' B' B (Fig. 65) a half-end pattern exactly as 
the half-end pattern E E' B' B in Fig. 64 is drawn, and with 
B' as centre and B' B as radius and B as centre and 6 h 



Fig. 65. 




(Fig. 62) as radiias, describe arcs intersecting in F. Join 
B F, F B' ; and produce F B' indefinitely. Make F P' equal 
to P B (Fig. 63), and with F' as centre and P C, P D and 
P B (Fig. 63) successively as radii describe arcs c, d, and e. 
With F as centre and h c (Fig. 63) as radius describe an aro 
cutting arc c in C, and with C as centra and same radius 
describe an arc cutting arc d in D. With same radius and 
D as centre describe an arc cutting arc e in E. Join 
GP', DP', EP'. Now with P' as centre and PC (Fig. 63) 



188 THE TINSMITHS' PATTERN MANUAL. 

as radius describe an arc c' cutting P' C in C, and witli earn© 
centre and P D', P E" (Fig, 63) succassively as radii describe 
arcs d' and e' respectively cutticg F D and P' E in D' and E'. 
Through P, C, D, and E draw an unbroten curved line. 
Also through B', C, D', and E' draw an unbroken cui-ved line. 
•Then B B F E E' B' B' will be the complete pattern required. 

Case III. — Pattern when the body is to be made up of one piece. 

In this case we will put the seam to correspond with B B' 
(Fig. 62). We now need an end pattern (the end & E 6 6' E' &' 
in plan), with right and left a side pattern attached [h Ah h', 
hDhh' in plan), and joined to each of these, a half-end 
pattern (b V B' B, 6 &' B' B in plan). 

First draw Fig. 63 ; then draw (Fig. 66) P E equal to PE 
(Fig. 63) and with P as centre and PD, PC and PB 
(Fig. 63) successively as radii describe arcs dd, cc, and h h. 
With E as centre and radius equal to "Ed (Fig. 68) describe 
arcs cutting arc d d right and left of P E in D and D. With 
points, D, D, successively as centres and same radius describe 
arcs cutting arc c c right and left of P E in C and G ; 
and with same radius and the last found points as centres 
describe arcs cutting arc h h right an|d left of P E in B and B. 
Join all the points found to P. With P as centre -and 
PE" (Fig. 63) as radius describe an arc cutting PE in E'. 
With same centre and PD' (Fig. 63) as radius describe an 
arc d! d' cutting lines P D right and left of P E in D' and 
D'. With same centre and P C (Fig. 63) as radius describe 
an arc c' c' cutting lines P G right and left of P E in 
C and C. Similarly by arc h'h' find points B' and B'. 
Through B, C, D, E, D, G, B, draw an unbroken curved line. 
Also through B', C, D', E', T>', C, B', draw an unbroken 
curved line. This gives us B E B B' E' B' a complete end 
pattern. Now with B' on thsT light-hand side of the 
end pattern as centre and B' B as radius, and B as centre and 
h h (Fig. 62) as radius describe arcs intersecting in F. Join 
B F, F B' ; produce F B' indeiSnitely, and to P B' attach the 
half-end pattern FE E' B' in precisely the same manner that 



THE TINSMITHS' PATTERN MANUAL. 



189 



F E E' B' the half-end pattern in Fig. 65 is attached to 
the side pattern B F B'. By a repe|,ition of tho foregoing 



Fig. 66. 





7\^ 

construction on the left of the end pattern B E B B' E' B' we 
can attach B B E.E' B' and complete E E E E' E' E' the pattern 
required. 

PEOBLEM XXIII. 

To draw, without long radii, the pattern for a tapering body 
with, top and base parallel, and having circular top and oblpng 
bottom with semicircular ends. The dimensions of the top and 
bottom of the body and its height being given. 

This problem is a fourth case of the preceding, and is 
exceedingly useful where the work ia so Isrge that it is 



190 



THE TINSMITHS' PATTERN MANUAL. 



incoBvenient to draw the whole of the plan, and to use long 
radii. 

To draw the pattern (with the body in four pieces, as in 
Case I. of preceding problem). 

(81.) -Draw (Fig. 67) E' E 6 A h' one quarter of the plan of 
the body. Divide the quadrants E 6, E' 6', each into the same 
nukber of equal parts, here three, in the points d, c, d', c' ; 



Fig. 67. 




join ddl c c'. Through E' draw E' F perpendicular to E' E 
and equal to the given height of the body. From E' along 
E' E mark off E' D, E C, and E' B respectively equal to d d\ 
c c', and h V ; and join E F, D F, C F, and B F ; then E F, 
D F, C F, and B F will be the true lengths of E S', d d', c c', 
and h V respectively. Next join d E', and draw E' e" perpen- 
dicular to it and equal to the given height, and join d e" ; then 
de" may be taken as the true length of d E'. Also join cd' 
and h a' ; through d' and c' draw cl' d" and c' c" perpendicular to 
c d' and h c' respectively, and each equal to the given hei.ght, 
and join c d" and h c" ; then c d" and h c" may be taken as the 
true lengths of d' c and c' 6 respectively. Through 6' draw 



THE TINSMITHS' PATTERN MANUAL. 



191 



V h" perpendicular to h' A and equal to the given heiglit, and 
join A h", then Ah" will be the true length of 6' A. 

Next draw (Fig, 68) E E' equal to E F (Fig. 67), and with 
E'- and E as centres and radii respectively equal to d e" and 
E d (Fig. 67) describe arcs intersecting in d, and with d and 
E' as centre and radii respectively equal to DF and E'd' 
(Fig. 67) describe arcs intersecting in d'. With d' and d as 
centres and radii respectively equal to c d" and d c (Fig. 67) 
describe arcs intersecting in c, and with c and d' as centres 

Fig. 68. 




and radii respectively equal to C F and d' c'(Fig. 67) describo 
arcs intersecting in c'. With c' and c as centres and radii 
respectively equal to h c" and c h (Fig. 67) d scribe arcs 
intersecting in 6. Similarly with 6 and c' as centres 
and radii respectively equal to B F and c' 6' (Fig. 67) describe 
arcs intersecting in h'. With h' and h as centres and radii 
respectively equal to h" A and b A (Fig. 67) describe arcs 
intersecting in A. Through E, d, c, b, draw an unbroken 
curved line. Also through E', d', c\ h\ draw an unbroken curved 
line. Join 6' A, b A"; then E c A 6' c' E' is the pattern required. 
(82.) The lines dS, cd, dW, &o., are drawn in Fig. 68 
simply to shov/ the position that the lines which correspond 
to them in Fig. 67 {dd\ cc', dE', &c.) take upon the developed 
surfat-e of the tapering body. 



192 



THE TINSMITHS' PATTERN MANUAL. 



PEOBLEM XXIV. 

To draw the pattern for a tapering hody with top and hase 
parallel, and having an oval bottom and circular top (^oval 
canister top, for instance). The height and dimensions of the 
top' and bottom of the body being given. 

Again four cases will be treated of; three ip this problem 
and one in the problem following (see also § 79, p. 157). 

Draw (Fig. 69) the plan of the body (see Problem IX. > 
p. 132), preserving of its construction the centres 0, O', P, P, 
and the four points (d) where the end and side curves of the 

Fig 60. 




plan of the bottom meet one another.ialso the points A'E' where 
the axis A E cuts the circular top. Join d A' (two places) and 
d E' (two places). From the plan we know (see g, p. 129) 
that dGdA'G'E',dBdA' B' E', the ends of the body, are like 
portions of the frustum of an oblique cone ; we also know 
that dAdA'jdEd E', the sides of the body, are like portion^ 
of the frustum of an oblique cone. 



TIIP] TINSMITHS' PATTERN MANUAL. 193 

(83.) It is evident tliat in tliis problem the arcs dG d, 
E'G'A' and dBd, E' B' A' are, neither • pair, proportional 
(§ 67, p. 124). We have hitherto in Problems XVIII., XX., 
and XXII. been dealing with proportional arcs. The work- 
ing will therefore differ, though but slightly, from that of 
problems mentioned. 

In Plate II. (p.-203) is a representation of the tapering body 
""for which patterns are required, also of two oblique cones 
(x and Z). The oblique cones show to what portions of 
their • surfaces the several portions of the tapering body 
correspond. Thus the sides, B', of the body correspond to 
the B portion of cone Z, and the ends, A', correspond to the 
portion A of cone x. The correspondence will be more fully 
recognised as we proceed with the problem. 

Case I. — Pattern when the body is to be made up of four 
pieces. 

We will suppose the seams are to correspond vnth. the 
plan lines G G', B B', A A', E E', of ends and sides, as in 
Problem XXII. .Then one pattern only, consisting of a 
half-end- pattern, with, attached, a half-side pattern will be 
required. 

To draw the pattern. 

Draw (Fig.'TO) separately G' Gfd A' the G' Gfd A' portion 
of Fig. 69, thus. Draw any line X X and with any point O 
in it (corresponding to 0, Fig. 69) as centre and G 
(Fig. 69) as radius describe an arc Gd, equal to G d oi 
-Fig. G9. Make G 0' equal to G 0' (Fig. 69), and with O' as 
centre and O'G' (Fig. 69) as radius describe an arc G'A' 
equal to G' A' of Fig. 69. Joining d A' completes the portion 
of Fig. 69 required Now divide GcZ.into any number of 
equal parts, here three, in the points /and e. From G' and" 
0' draw G' G", 0' 0" perpendicular to X X and each equal to 
the given height of the body. Join G G"„ 0" ; produce 
them to inter.-ect in S (§ 80, p. 158) ; from S let fall S S' per- 
pendicular to X X, and join 0" G". Now join d to S', by a 
line cutting arc G'A' in d\ then (§ 68, p. 12i, and d, p. 126) 



194 



THE TINSMITHS' PATTERN MANUAL. 



d! and d are corresponding points and (§ 67, p. 124) tbe 
arcs Gr d, G' d' are proportional ; G G' d' £? is the plan of a 
portion of an oblique-cone frustnm, lying between the same 
generating lines, and d d' A' is plan of a portion of the same 
frustum outside the generating lino S' d. With S' as centre 

Fig, 70. 




and S'/, S' e, and S' d successively as radii describe arcs 
cutting X X in F, E, and D ; join these points to S by lines 
cutting 0"G" in F', E', and D', With S' as centre and 
radius the distance between S' and A' describe an arc A' A 
cutting X X in A ; from A draw A A" perpendicular to X X 
and cutting O" G" in A", and join A" S. 

Next draw any line X X (Fig. 71) and with any point P 
in it (corresponding to P, Fig. 69) as centre and radius PA 
(the radius of arc d A.d., Fig. 69) describe an arc A d equal to 
Ad of Fig, 69. Make A A' and A 0' respectively equal to 



THE TINSMITHS' PATTERN MANUAfc, 



195 



A A' and A 0' (Fig. 69). Divide A d into any nurater of 
equal parts, here three, in the points 6 and c. Prom A and 
0' draw A' A", O' O" perpendicular to X X and each equal to 
the given height of the body. Join A A", PO"i produce 

Fig. 71. 




them to intersect in Q (§ 80, p. 158) ; from Q let fall Q Q' per- 
peadio'llar to X X, and Join O" A". With Q' as oentte and 
Q' &, Q' c, and Q' d successively as radii describe arcs cutting 
XX in B, C, and D, and join these points to Q. 

Nest draw (Fig. 72) a lino S G equal to S G (Fig. 70) and 
ifirith S 843 centre and S F, S E, and S D (Fig. 70) successively 
as radii describe arcs /, e, and d. With G as centre and 
radius equal to G/ (Fig. 70) describe an arc cutting arc / in 
F. With same radius and F'as centre describe an arc cutting 
arc e in E, and with S aa centre and same radius describe an 
arc cutting arc d in D. Joia th© points F, E, and D to S. 
With S as centre and S G" (Fig. 70) as radius describe an 
arc g' cutting S G in G'. With samo centre and S F' (Fig. 70) 



196- 



THE TINSMITHS' PATTERN MANUAL. 



as radius describe arc /' cutting B P in F'. With same centra 
and S E' (Fig. 70) as radius describe an arc e' cutting S E in 
E', and with same centre and S D' (Figo 70) as radius describe 
arc d' cutting S D in D'. With same centre and S A" (Fig. 70) 
as radius describe arc a', and with D' as centre aifd radius d! A' 
(Fig. 70) describe an are intersecting arc a' in A'. Make 
D Q equal to D Q (Fig. 71) and with Q as centre and Q C, 

Fig. 72. ; 




Q B, and Q A (Fig. 71) successively as radii describe arcs c, 6, 
and a. With D as centre and d c (Fig. 71) as radii describe an 
a,rc cutting arc c in C, and with C as centre and same radius 
describe an arc cutting arc b in B. Similarly with same 
radius and B as centre find point A. Join A A'. Through 
the points G, F, E, D, 0, B, A draw an unbroken curved 
line. Also through the points G', F', E', D', A' draw an un- 
broken curved line. Then "GD A A'E'G' is the pattern 
required.' 



1}HiE TINSMITHS' PATTERN MANUAL. 



197 



Case II. — Pattern -when tlie "body is to he made np of two 

pieces. 
We will suppose tlie seams are to correspond witli tlie lines 
G G' and B B'. It is evident tliat here we need but one 
pattern only, wHda will combine a side of the body and two 
half-ends, in fact will be double that of Pig. 72, 

First draw B D A'B' (Fig. 73) a half-end pattern exactly 
as the half-end pattern G D A' G' in Fig. 72 is drawn, and 
make DQ equal to D Q (Fig. 71). With Q as centre and 

Fm. 73. 




QC, QB, and Q A (Fig. 71) successively as radii describe 
arcs dd, c c, bb, and a a. With D as centre and radius equal 
to dc (Fig. 71) describe an arc cutting arc cc in C, and with 
same radius and C as centre describe are cutting arc 6 6 in B. 
With B as centre and same radius describe an arc cutting are 
a in A, and with same radius and A as .centre describe an 
arc cutting arc bb in B, Similarly with same radius and 



lOS THE TINSMITHS' PATTERN MAXtJAL. 

points B and C, to the right of A, successively as centres find 
points C and D. Join D Q and produce it indefinitely ; make 
D S' equal to S D (Fig. 70). With S' as centi-e and S E, S F, 
and S G (Fig. 70) successively as radii describe arcs e, f, and 
g, and with D as centre and d e (Fig. 70) as radius describe 
an arc cutting 'arc e in E. With same radius and E as centre 
describe an arc cutting arc / in F. Similarly with F as 
centre and same radius find point G. Join the points E, F, 
and G to S'. With S' as centre and S D' (Fig. 70) describe 
arc ff cutting S' D in D'. With game centre and S E' (Fig. 
70) as radius describe arc e' cutting S' E in E'. Similarly 
with same centre and S F' and S G" (Fig. 70) successively as 
radii find points F' and G'. Through tho-points D, C, B, A, 
B, C, D, E, F, G, draw an unbroken curved line. ALso 
through points A', D', E', F', G' draw an unbroken curved 
line. Then B A G G' A B' is the complete pattern required. 

Case III. — Pattern when the body is to be made up of erne 

piece. 

In this case we will put the seam to correspond with G G' 
(Fig. 69). We now need an end pattern (the end dBdA 
B' E' in plan), with right and left a side pattern attached 
{dAdA', t? E d E' in plan), and joined to each of these a half- 
end pattern {d A' G' G, d E' G' G in plan). 

Fircit draw Figs. 70 and 71 ; then draw (Fig. 74) S G equal 
to S G (Fig. 70), and with S as centre and S F, S E, and S D 
(Fig. 70) STiCcessively as radii describe arcs//, ee, and dd. 
With G as centre and radius G/ describe arcs cutting arc// 
right and left of S G in F and P. With points F, F, succes- 
sively as centres and same radiias describe arcs enttingaro e0 
right and left of S G in E and E, and with same, radius, 
and the last found points as centres describe arcs cutting 
arc d d right and left jof S G in D and B. Join all the points 
found to S. With S as centre and S G" (Fig, 70) as radius 
dt scribe an arc cutting SG in G\ With same centre and 
S F' (Fig. 70) as radius describe an arc /'/' catting lines 
SF right and -left of SG in F' and F. With same centra 



METAL-PLATE WORK. 



199 



and S E' (Fig. 70) describe an arc e' e' cutting lines S E 
right and left of S G in E' and E', and with same centre and 
S D' (Fig. 70) as radius describe arc d! d' cutting lines S D 
right and left of S G in D' and D'. With S as centre and D' 
right and left of S G as centres and radii respectively equal 
to S A" and d' A' (Fig. 70) describe arcs intersecting in A' 
and E'. Through D, E, F, G, F, E, D, draw an unbroken 
(Mrvedline. Also through E', D', E', F', G', F', E', D', A', draw 




a:a unbrolcen curved line, and join D E', D A'. This gives us 
D G D A' G' E' a complete end pattern. Now attach the side 
pattern D A D A' and the half-end pattern D G G' A' to the 
right and left of the complete end pattern we started with, 
in precisely the same manner that the side pattern D A D A' 
and half-end pattern D G G' A' in Fig. 73 is attached to D A', 
which corresponds to D A' in Fig. 74. This will complete 
G E G A G G' G' G' the pattern required. ,. 



200 



THE TINSMITHS' PATTERN MANUAL. 



PROBLEM XXV. 

To draw, without long radii, the pattern for a tapering ho^y 
with top and base parallel, and having an oval bottom and 
circular top. The height and dimensions of the top and base 
of the body being given. 

This problem is a foy.rtli case of tlio preceding, and is 
exceedingly useful where the work is so large that it is- 
inconvenient to draw the whole of the plan, and to use long 
radii. 

T)3 draw the pattern (with the body in four pieces, as in 
Case I, of preceding problem). 

(84.) Draw (Fig. 75) E (^ & A A'-d' E' one quarter of the plan 
of the body. Join c (the point where the end and side curves 



Fig. 7S. 




of the plan of the bottom meet) to A', the extremity of the 
uadrant E A'. We must now get at corresponding points in 



l-HE TINSMITHS' PATTERN MANUALf 20l 

the arcs E c, E' A'. To do thio, as the arcs are not propor- 
tional (§ 67), we must find the plan of the apex of the 
oblique cone of a portion of which E c A' E' is the plan. It 
is in the finding of these points that our present working 
differs from the working of Problems XIX., XXI., XXIlI.,and 
XXVII., where the corresponding points d d', cc' (Figs. 50, 58, 
60, and 82) are found. With radius O E produce arc E c inde- 
finitely, and through draw Q perpendicular to E'E and 
cutting E c produced in Q. Then E Q is a quadrant, and E Q, 
E' A' (each a quadrant) are proportional. In Q and A' therefore 
we have corresponding points (§ 68), as well as in E and E', 
which are points on the longest generating line (6, p. 126) ; and 
the intersection of O 0' produced, of which E E' is part, and 
Q A' joined and produced will give us V, the required plan 
of the apex. Next divide E c into any number of equal parts, 
here two, in point d. Join dY, cY cutting E' A' in d' and c', 
respectively (the lines from d and c are not carried to V in 
the fig.)s thtn (^ and d', c and c' are corresponding points, 

Kext through E' draw E' F perpendicular to E' E and 
equal to the given height; from E' along E'E mark off 
E' D, E' C, respectively equal to d d' and c c', and join E F, 
DF, and CF; then EF, DF and CF will be the true 
lengths of E E', d d', and c c'. Join E d' and d c' ; through d' 
and c' draw d' d" and c' c" perpendicular to d' E and c' d re- 
spectively, and each equal to the given height, and join 
E d", t? c" ; then E(i"and dc" may be taken respectively as 
the true lengths of d' E and c' d. Now divide arc A c into 
any number of equal pans, her© two, in the point b, and join 
6 A'; through A' draw three lines A' A"; one perpendicular 
to A' c, the second perpendicular to A' 6, and the third per- 
pendicular to A' A, and each equal to the given height, and 
join c A", h A", and A A" ; then c A", 6 A", and A A" may ha 
taken as the true lengths of A' c, A' b, and A' A respectively. 

Nest draw .(Fig. 76) E E' equal to E P (Fig. 75), and with 
E and E' as centres and radii respectively equal to E d" and 
E' d' (^Fig- 75 j de.«cribe arcs intersecting in d\ and with d' and 
E as centres and radii respectfully equal to DF and Ed 



'm 



^HE TINSMITHS* PATTERN MANUAL. 



(Fig. 75) describe arcs intersecting in- d. With i an^ 3! as 
centres and radii respectively equal to dc" and d! c' (Fig. 75) 
describe arcs intersecting in c', and with c' and d as centres 
and radii respectively equal to OF and dc (Fig. 76) de- 
scribe arcs intersecting in c. With c and c' as centres and 
radii respectively equal to cA"and c'A' (Fig. 75V describe 

Fis. 76. 




arcs intersecting in A'; and with A' and c as centres and radii 
h A" and c b (Fig. 75) describe arcs intersecting in &. Simi- 
larly with A' and b as centres and radii respectively equal to 
A A" and 6 A (Fig. 75) describe arcs intersecting in A. 
Juin A A'. Through E, d, c, b, A draw an unbroken curved 
line. Also through E', d', c\ A' draw an unbroken curved 
line. Then E c A A' E' is the pattern required. 

The lines d d', c c', E d', d c', &c., are not needed for the 
working, they are drawn for the reason stated in § 82, end 
of Problem XXHI. 

(85.) If Y is inaccessible, corresponding points e, c'. d, d' 
can thus be found. From the point E' along the arc E' A' 
set off an arc proportional to the arc E c in. the following 
manner. Join O c (line not ^own in fig.) and through O' 
draw O'c* (also not shown in fig.) parallel to Oc and 
cutting arc E' A' in- c' ; then arcs E' c' and E c will be pro- 
portional. (The student must particularly notice this 



THE TINSMITHS' PATTERN M\NUAL, 



205 



metliod of drawing proportional arcs. It is outside the scope 
of the book to prove the method.) Now divide arcs E' c\ 
E c each into the same number of equal parts, here two, in 
the points d and d' ; then (^, (^' and c, c' are corresponding 
points.. 

PROBLEM XXVL 

To draw the pattern for a tapering body witli top and base 
parallel, and having oblong bottom with round (quadrant) 
corners, and circular top. The dimensions of the top and 
base of the body and its height being given. 

Again four cases will be treated of, three in this problem, 
and one in the problem following so § 79, p. 157). 

Draw (Fig. 77) the plan of the body (see Problem X., 
p. 133) preserving of its construction the centres 0' and the 

Fig. 77. 




points E' B' where the flat sides and flat ends meet the circle 
of the top. Join B B', E E' each in four places. From the 
plan we know (see g, p. 129) that the round corners of the 
body are portions of frusCa of oblique cones, 



206 THE TINSMITHS' PATTERN MANUAL. 

(86.) Looking at the plan, we can at once see that what 
we have to deal with diffors somewhat from what has as yet 
been before us. Hitherto a line passing through the centres 
of the plan arcs bisected the arcs, and the cone development 
was consequently identical each side of a central line. In 
Fig. 77, however, the line drawn through O O' does not 
bisect the plan arcs E B, E' B'. This affects the working 
but little, as will be seen. 

In Plate III. (p* 213) the tapering body is represented ; 
also an oblique cone Z, the A portion of which corresponds 
to the A' portion of the body, and the development of the 
former is the development of the latter. 

Cass I.— Pattern when the l)ody is to be made up of four 

pieces. 

We will suppose the seams to correspond with ihe plan 
lines G E', D E', F B', A B', of ends and sides, as in Problems 
XXII. and XXIV. just preceding. Then one pattern, com- 
prising a half-end, a complete corner, and a half-side, will 
be the pattern required. 

To draw the pattern. 

Draw separately (Fig. 78) an E E', B' B portion of Fig. 77, 
thus. Draw an indefinite line S' d (Fig. 78), and with any 
point O (corresponding to O, Fig. 77) in it as centre and B 
(Fig. 77) as radius describe an arc B E. Join O O' (Fig. 77) 
and produce it cutting arc B E in d ; make d B and d E (Fig. 
78) equal respectively to <i B and d E (Fig. 77). Now (Fig. 
78) make d O' equal to e? O' (Fig. 77), and with 0' as centre and 
0' B' (Fig. 77) as radius describe an arc B' B'. - Make d' B' and 
d'E' equal respectively to d'Jj' and d'E' (Fig. 77). Joining E B', 
BB' completes the portion of Fig. 77 required. S'ow divide 
(Fig. 77) B E into any number of parts. It is convenient to 
take d as one of the division points, and to make d c equal 
to d'E; leaving cB without further division, thus making 
the division of B E into three portions not all equal. In 
actual practice the dimensions of the work will suggest the 
number of parts espedieat. Now (Fig. 78} make d c equal 



THE TIXSMITIl.-^' PATTERN MANUAL. 



207 



to d c (Pig. 77), then B E (Fig. 78) will be divided corre- 
Bpondingly to BE (Fig. 77). Draw XX parallel to S'<?; 
and at d and O draw d D, O Q perpendicular to 8' d; and 
meeting X X in D and Q ; also through d' and O' draw d' B', 



Fio. 78. 




O' 0" perpendicular to S' «l j the line O' O" cutting X X in Q'. 
Make Q' O" equal to the given height of the body, and draw 
O" D' parallel to X X. Join D D', Q 0" ; produce them to 
intersect in S (§ 80, p. 158) ; and from S let fall a perpsn- 
aiciilar to S' J, cutting B' d in S'. With S' as centre and S' S, 



208 



THK TINSMITFIS- PATTERN MANUAL. 



or S' c (whicli is equal to S E) and S' B successively as radii 
describe arcs cutting S' d in e and/. Draw e C, /F per])en- 
dicular to X X and cutting it in G and F ; also join C S, F S, 
cutting O" D' in C and F'. 

Next draw 8 D (Fig. 79) equal to S D (Fig. 78), and with 
S as centre and SO, SF (Fig. 78) successively as radii 

Fig. 79. 




arcs c and b. With D as centre and radius equal 
to d'E (Fig. 78) describe arcs cutting arc cc in E and C. 
Witli C as centre and radius cB (Fig. 78) describe an arc 



THE TINSMITHS' PATTERN MANUAL, 209 

cntbing arc h in B. Join E, C, and B to S. Make S D' equal to 
S D' (Fig. 78) and with S as centre and S C (Fig. 78) as radius 
describe an arc c' c' cutting S E and S C in E' and C respec- 
tively. With same centre and S F' (Fig. 78) as radius describe 
an arc V cutting S B in B'. Through E, D, C, and B draw an 
unbroken curved line. Also through E', D', C, and B' draw 
an unbroken curved line ; this will complete the pattern of 
the round corner. To attach the half-end and half-side patternw 
to E E' and B B' respectively, the true lengths of E' D and 
jB' A (Fig. 77) must be found. Draw (Fig. 77) E' H perpen- 
dicular to E' D and equal to the given height of the body ; 
join D H, then .D H is the true length of E' D. The lines 
B' E and B' A being equal, their true lengths are equal, we 
will therefore for convenience find the true, length of B'A 
in that of B' F. Draw B' G perpendicular to B' F and equal 
to the given height, join F G, then F G is the true length 
required. Now with E' (Fig. 79) as centre and D H 
(Fig. 77) as radius, and E as centre and radius ED (Fig. 77) 
describe arcs intersecting in G. Join EG, G E' ; this 
attaches to E' E the half-end pattern. With B' (Fig. 79) 
as centre and F G (Fig. 77) as radius, and B as centre and 
radius B A (Fig. 77) describe arcs intersecting in A. Join 
B A, A B' ; this attaches to B' B the half-side pattern. Then 
A B' E' G is the complete pattern required. 

Case IL^— Pattern when the body is to be made up of two 

pieces. 

Here it will be best that the seams shall correspond with 
the liijes AB', FB', ihatis with the middle of each side. 
The required pattern will then be double that of Case I. 

Draw (Fig. 80) EBB'E', the corner pattern, in exactly 
the same manner that E B B' E' (Fig. 79) is drawn. With 
E' as centre and E' E as radius, and E as centre and E B 
(Fig. 77) as radius describe arcs intersecting in F. Join 
E F, F E'. Broduco F E' indefinitely and make F S' equal to 
ES. Using S' as centre, the round corner FGB'E' can be 
drawn as was E B B' E. With B' as centre and F G (Fig. 77) 



210 



THE TINSMITHS' PATTERN MA^TFAL. 



as' radius, and B as centre and B A. (Fig- 77) as radius de- 
scribe arcs intersecting in A. Similarly with B' and G aa 



Fia. 80. 




centres and same radii respectively describe arcs intersecting 
in F. Join B A, A B', G F, F B', then A D F P B' E' B' will 
be the pattern required. 

Case III.— Pattern when the body ia to bo made up of one 

piece. 

We will suppose the seam to correspond with C E' the 
middle of one end. Draw G F D B B' E' B' (Fig. 81) in th© 



i'HE TINSMITHS' PATTERN MANUAL. 



211 



same manner that G F D B B' E' B' (Fig. 80) is drawn. With 
B' as centre and B' B as radius, and B as centre and B B 
(Fig. 77) as radius describe arcs intersecting in A. Join 
B A, A B' ; produce A B' indefinitely. Make A P equal to 
S F (Fig. 78). With P as centre and S C and S D succes- 




eivoly as radii describe arcs h and I, and with A as centre 
and B c (Fig. 78) as radius describe an arc cutting arc h in 
H. With H as centre and c d (Fig. 78) as radius describe an 
»rc cutting arc Hn L ; and with L as centre and same radius 



212 



THE TINSMITHS' PATTERN MANUAL. 



describe an arc cutting arc li in M. Join H, L, and M to P, 
With P as centre and S C (Fig. 78) as radius describe arc h' 
cutting P H and P M in H' and " M' respectively. With 
same centre and SD' (fig. 78) as radius describe arc V 
cutting PL in L'. Through the points A, H, L, M draw an 
unbroken curved line. Also through the points B', H', L', M' 
draw an unbroken curved line. With M' as centre and D H 
(Fig. 77) as radius, and M as centre and E C (Fig. 77) as 
radius describe arcs intersecting in N. Join M N, N M'. 
Eepeating the working to the left of B'G, the GOO'B' 
portion of the pattern can be drawn which completes the 
pattern required. 

PEOBLEM XXVII. 

To draw, without long radii, the pattern for a tapering body 
with top and base parallel, and having oblong bottom with 
round quadrant corners, and circular top. The dimensions 
of the top and base of the body and its height being given. 

This problem is a fourth case of the preceding, and is 
exceedingly useful where the work is so large ihat it is in- 
convenient to draw the whole of the plan, and to use long 
radii. 

To draw the pattern^ (with the body in four pieces as in 
Case I. of preceding problem). 

Draw (Fig. 82) Ecb Ab'c' d', one quarter of the plan of 
the body. Divide the arc (quadrant) d 6 into any number 
of equal parts, here two, in the point c, and the arc d' b' into 
the same number of equal parts in the point c' ; and join cc'. 
Through d' draw d' F perpendicular to d' E and equal to the 
given height of the body. From d' along d' E mark off d' B, 
equal to 6 6', and d'D equal to cc' and dd' (which two lines 
have happened to come in this particular fig. so nearly 
equal that we may take them as equal), and join B F, E F, 
and D F ; then B F and E F will be the true lengths of 6 b' 
and l^d' respectively, ^nd DF may be taken as the true 



THE TINSMITHS' PATTERN MANUAL. 



nr> 



length of both cc' and dd'. Next join c'd,h'c; draw c' c", 
b' h" perpendicular to c' d, V c respectively, and each equal to 
the given height ; and join d c" and c h" ; then d c", c h" may 



Pio. 82. 







(f 


d 


h 


V 












BE 








cZ. 




„^_^^ 






b' 




"^"^ ^ 


~-^<f 


6'X 


«) ^ 


4 


"t 


^ 


^ 



he taken as the tme lengths of c' d and h'e respectively. 
Draw b' b" perpendicular to 6' A and equal to the given 
height, and join A 6", then A b" is the true length of 6' A. 

Next draw (Fig. 83) d d' equal to D F (Fig, 82), and with 
d and d' as centres and radii r^pectively equal to d c" and 
d' g' (Fig. 82), describe arcs intersecting in c'. With c' and 
d as centres, and radii respectively equal to D F and d c 
fFig. 82) describe arcs intersecting in c'; and with c and c' 
as centres and radii respectively e<]ual c b" and c' 6' (Fig, 82) 
describe arcs intersecting in b' ; also with b' and c as centres 
and radii respectively equal to B F and ch (Fig. 82) describe 
arcs intersecting in &. With d' and d as centres and radii 
respectively equal to E F and d'E (Fig. 82) desciibe arcs 
intersecting in E ; and with b' and b as centres and radii 
rcsj^ieutively equal to b" A and 6 A (Fig. 82) describe area 



21 (j THE TINSMITHS' PATTSUN MANUAL, 

intersecting in A. Throngli d, c, h, draw an unbroken curved 
line. Also through d', c', h' draw an unbroken curved lino. 




Join d E, d' E, h A, and h' A ; then E c A J'tT is the pattern 
required. 

The lines c c', h h', &c.', are not needed for the working, 
they are drawn fur the reason stated in § 82, end of Problem 
XXIII. ' 



PROBLEM XXVIII. 

To draw the pattern for an Oxford hip-bath, the like dimensions 
to those for Problem XI. being given. 

It is only necessary to treat of two cases, one m this 
problem, and one in the problem following (see also § 79, 
p. 157). 

Draw (Fig. 84) the plan of the body (see Problem XI., p. 134), 
preserving of its construction the centres O, P', P, Q', Q ; the 
points D, D' (two sets) in which the arcs, in plan, of the 
back and sides meet each other ; and the points h, y' (two 
sets) in which the plan arcs of the sides and front meet each 
other. Join h g' (two places) as shown in the fig. Esamining 
the plan of the bath we see (d, p. 55) that the back of it, 



•THE TINSMITHS' PATTERN MANUAL. 



217 



D A D D' A' D', is a portion of a right cone ; that the sides 
DD' g'h are {g, p. 129) each of them a portion of an oblique 
cone; and that the portion hg'g'h is also a portion of an 




oblique cone.' Similarly as in Problem XXIV. (§ 83, p. 193) 
the arcs D h, D' g' and B A, B' g' are, neither pair, propor- 
tional. 

In Plate IV. (p. 227) is a representation of an Oxford hip- 
bath, also of a right cone x, and two oblique cones Y and Z. 
The cones show to what portions of their surfaces the several 
portions of the bath correspond. Thus the back. A', of the 
bath corresponds to the A portion of right cone x ; the half- 
fronts, C, of the bath correspond each of them to the C 
portion of oblique cone Y, and the sides, B', of the bath 
correspond each of them to the B portion of oblique cone Z. 

Patterns when the body is to be made up of three 
pieces. 

We wili put the seams tc correspond with the linos DD' 



218 I'HE TINSMITHS' PATTERN MANUAL. 

(two places), and B B', Clearly, only two patterns will be 
required, one for tlie back of the bath, and the other for 
a complete side and a half-front. 

To draw the pattern for the back. 

Draw E A A'0'0 (Fig. 85) the elevation of the back (see 



Fig. 85. 









Problem XI. and Fig. 24a, p. 135) and produce A A', 0' to 
intersect in O". With O as centre and A as radius describe a 
quadrant A D (corresponding with A D, Fig. 84), and divide 
it into any number of equal parts, here three, in the points 
b and c. Draw c C, 6 B perpendicular each of them to A O 
and cutting it in points B and C. Join O" C, and produce it 
to cut E in C ; and through C draw C C" parallel to A O 
and cutting O' E in C". Join O" B and produce it to cut 
O E in B' ; and through B' draw B' B" parallel to A O and 
cutting O" E in B". Then A C" is the true length of C C, 
and A B" the true length of B B', ' 



THE TINSMITHS' PATTERN MANUAL. 



219 



Nest draw (Fig. 86) O" A equal to^O"A (Fig. 85), and 
with 0" as centre and radius O" A describe an arc DAD, and 
with same centre and 0"A' (Fig. 85) as radius describe an 







^^\uit/ 



arc D' A' F. Mark off, right and left of A, A S, B C, and 
C D, each equal to A 6 (Fig. 85), one of the equal parts into 
which the quadrant A D is, divided. Join D O", C O", B O'' 
right and left of A O", and produce 0" A, O" B, 0" C indefi- 
nitely. Make A E equal to A E (Fig. 85) ; B B" equal to 
AB" (Fig. 85); and CC". equal to AG" (Fig. 85); and 
through D, G", B", E, B", C", D, draw an unbroken ciirved 
line. Then D E D D' A' D' is the required' pattern for back 
of bath. 

To draw the pattern for a side and a half-front. 



THE TINSMITHS' PATTERN MANUAL. 



Draw separately D' D fJi g' (Fig. 87), the B'Tiflig' portion 
of Fig. 84, thus. Dra,w any line X X and with any point Q 
(to correspond with Q, Fig. 84) in it as centre and (same 

Fia 8'?, 




fig.) Q D (tha D on the further side of A B) as radiws' describe 
an arc D h equal to D fe of Fig. 84. Mat^ D Q' equal to D Q' 
(Fig. 84), and with Q' as centre and Q' D' (the farther D', 
Fig. 84.) as radius descriho an arc D'^' equal to D'^' of Fig. 84, 



THE TINSMITHS' PATTERN MANUAL. 



221 



Joining h g' completes tlie portion of Fig. 84 required. Now 
from D' and Q' draw D' D", Q' Q" perpendicular to X X and 
each equal to the given height of the D B D portion of the 
bath. Join D D", Q Q" ; produce them to intersect in S 
(§ 80, p. 158); and from Slet fall S S' perpendicular to XX. 
Join S' g', and produce it to cut arc D h in g, then g and g' 
will be corresponding points. Divide D g into any number of 
equal parts, here three, in the points e and/. Join Q" D". With 
S' as centre and S' e, S'/, and S' g successively as radii describe 
arcs cutting X X in E, F, and G ; join these points to S by 
lines cutting Q" D", produced, in E', F', and G'. With S' as 
centre and S' Ji as radius describe an arc cutting X X in H, 
and join H S. 

Next draw separately B' B ^ «;' (Fig. 88) one ■ of the 
B'Bhg' portions of Fig. 84, thus. Draw any line X X and 




with any point P (to correspond with P, Fig. 84) in it as 
centre and P B (Fig, 84) as radius describe an arc B h equal 
to B li of Fig. 84, Make B P' equal to B P' (Fig, 84), and 
with P' aS centre and 1 ' B' (Fig. 84) as radius describe an arc 



222 



•niK Tl.VSMfniS- PATTERN MANUAL. 



IV g' equal to B' g' of Fig. 84, Joining h g' completes tlie por- 
tion of Fig. 81 required. Now from B' and F draw B' B", 
F' P" perpendicular to X X and cacli equal to the given 
hciglit of the D B D portion of the bath. Join B B", P P" ; 
prodiice tlieia to intersect in R (§ 80, p. 158) ; and from K 
lot fall li K' per])endicular to X X. Join R' h, cutting arc 
B' g' in h', then h and h' will he corresponding points. 
Divide B^ into any number of equal parts, here two, in the 
pointy. Join P" B". "With E' as centre and B! j and E'A 
successively as radii describe arcs cutting XX in J and H; 
join these points to B by lines cutting P" B" in J' and 11'. 
Next draw (Fig. 89) a Hno D S equal to D S (Fig. 87). 



FiQ. 89. 




With S as centre and S E, S F, S G, and S U (Fig. 87) sticces- 
eiyely as rs^dii describe arcs e, /, g^ and ft. Witli D as centro 



THE TINSMITHS' PATTERN MANUAL. 223 

and radius equal to D e (Fig. 87) describe an arc cutting arc 
e in E, and with same radius and E as centre describe an arc 
cutting arc / in F. With F as centre and same radius de- 
scribe an arc cutting arc g in G, and with G as centre and 
radius g h (Fig. 87) describe an arc cutting arc h in H. Join 
the points E, F, and G (not H) to S. Make S D' equal to 
S D" (Fig. 87) ; and make S E', S F', and S G' respectively 
equal to S E', S F', and S G' (Fig. 87). Through the points 
D, E, F, G, H draw an unbroken curved line. Also through 
points D', E', Fj G'tlraw an unbroken curved line, and join 
H G'. This completes the side pattern, to which we have 
now to attach, at H G', a half-front pattern. With H and G' 
as centres and radii respectively equal to H H' and g' h' 
(Fig. 88) describe arcs intersecting in H'. Join H H' ; pro- 
duce it indefinitely and make H R equal to H R (Fig. 88). 
With R as centre and R J, R B successively as radii describe 
arcs/ and 6. With H as centre and kj (Fig. 88) as radius 
describe an arc cutting arc j in J, and with same radius and 
J as centre describe an arc cutting arc h in B. Join the 
points J and B to R, and make R J' and R B' respectively 
equal to R J' and R B" (Fig. 88). Through H, J, and B draw 
an unbrokeu curved line. Also through H*, J', and B' draw 
an unbroken curved line. Then DFHBB'G'D' is the 
compiet© pattern rec^tdred, 



X 

To drawj withotit long radii, the patiem for an Oxford 
Mp-hatTi ; given dimensions as before. 

This problem is a second caso of the preceding. 

Patterns when the body is to be made up of three pieces 
with seams as in the preceding problem. 

To draw the pattern for the back. 

Draw E A A' 0' O (Fig! 90) the ©levEtion of the back as ia 
Fig. 86, and producs 0' 0. With as oeatr© and A as 



:4 



THE TINSMITHS- PATTERN MANUAL. 



radius describe a quadrant A D (^corresponding witli A D, 
Fig. 84), and divide it into any number of equal parts, here 
three, in the points b and c. Join & O, c O, and with O as 
centre and radius O'A' describe a quadrant D'A" (corre- 
sponding with D' A', Fig. 84), and cutting lines O c, O 6 in 
c'' and' b' respectively. Then D D' A" A will bo the plan of 
that portion of the back of the bath of which O' A'O A is the 
elevation. Through 6 and c draw &B and c C perpendicular 
to O A ; and through b' and c' draw b' B' and c' C perpendi- 
cular to 0' A'. (Here part of b' B' happens to coincide with 




A B' C 0' 



part of c C). Join B' B, C C, and prodnco them to meet E 
in B" and C" respectively. Through C" draw 0" c" parallel 
to A, and through B" draw B" b" parallel to A, Through 
D' draw D' D" perpendicular to D' D and eqnal to the given 
height of that portion of the bath, and join D D", then D D" 
will be the true length of B' D. Join D c' ; through c' draw 
c'F perpendicuLir to c'D and equal to the given height, and 
join D F, then D F may be taken as the true length of D c'. 

In drawing the pattern, we will first set out that for the 
O A O' A' portion of the back, and then attaoh to it the 
pattern for the O E A portion. It is evident that the A 0' A' 



THE TINSMITHS' PATTERN MANUAL. 



225 



portion is half a right cone frnstum, and therefore its pattern 
can be drawn (see Problem VIII., p. 41), thus. Draw (Fig, 91) 
a line D D' (the line D D' left of E A') eciual to D D" (Fig. 
90). With D and D' as centres and radii respectively equal 
to D F and D' c' (Fig. 90) describe arcs intersecting in c'. 
With D' and D as centres and radii respectively equal to 
D F and D c (Fig. 90) describe arcs intersecting in c. To 
'xind points 6 and b' proceed as just explained and with the 
same radii, but c and c' as centres. in!stcad of D and D', 




Similarly to find A and A', and the points 5 and 6', &c., on the 
^right-hand side of A A'. Through the points D', c', b\ A', &', 
c', D' draw an unbroken curved line. Also through D, c, 6, 
A, 6, c, D draw an unbroken curved line, and join JD D'. This 
completes the pattern for the OAO'A' portion of back of 
bath, to which we have now to attach, at D A D, the pattern 
for the O E A portion. Join A' A ; produce it indefinitely 
and make A E equal to A E (Fig. 90). Next join, right and 
left of A A', 6' 6, c' e, and produce them indefinitely; make 
b b", right and left of A A', equal to A b" (Fig. 90). Also 
make c c'\ right and left of A A', equal to A c" (Fig. 90}, and 



226 



THE TINSMITHS' PATTERN MANUAL. 



tlirough tlie points D, c", 6", E, b", c", D draw an nnhrokeu 
curved line ; then D E D D' A' D' is tbe complete pattern 
required. 

To draw tlie side and halfrfront pattern. 

Draw separately (Fig. 92) the BhDD' g' B' portion of 
Fig. 84, that is, the plan, of a side and half- front of the bath. 
Join h g' (as was done ia Fig. 84). We now have to get at 
corresponding points in the arcs D h, D' g\ and also in the arcs 
h B, ^' B . We do this by the method given in § 85, p. 202, 

EiG. 92. 




thus. Join Q' g' and throngh Q draw Q g parallel to Q' g* and 
cutting arc Dh in g; then the arcs D g and D' g' v,dli be the 
proportional. Now divide arc D g into any number of equal 
parts, here two only to avoid confusion, in the point e, and 
divide the arc D' g' into two equal parts in the point e' ; then 
s, e' and g, g' are corresponding points. Next join P h and 
through F draw F h' parallel to P ^ and cutting arc B' gr' in lt\ 
Pivide arcs B h and B' h' each into the same number of equal 



THE TINSMITHS' PATTERN MANUAL. 



229 



g;jrts, here two, in the points /and /' respectively; then/,/' 
and %, h' are corresponding points. Join e e', g g\ ff, and h h'. 
Through B' draw B' A perpendicular to B' B and equal to the 
given height of that portion of the bath; and from B' along 
B' B mark off B' F, B' H, B' G, and B' E respectively equal to 
//', h h', g g', and e e. Join B A, FA, HA, G A and E A ; 
then B A, FA, li A, G A, and E A will be respectively the 
true lengths of B B', //, h h\ g g', and' e e'. Join /' B, h' f, e' g, 
and D'e; through/' draw /'/" perpendicular to /'B, and 
equal to the given height, and join B/", then B/" may be 
taken as the true length of /' B. Through h', g', e', and D' 
draw h'h", g' g", e' e", and D'D", perpendicular to /»'/, g' h, e' g, 
and T)' e respectively, and each equal to the given height; 
also draw D' D" perpendicular to D D' and equal to the given 
height; and join fh". Jig" ge", eD", and DD"; then/A", 
hg", ge", eD", and D D" may be taken as the triie lengths 
oili' f, g'h, e' g, D'e, and D'D respectively. 



Fig. 93. 




Next draw (Fig. 93) B B' equal to B A (Fig. 92), ana with 
B and B' as centres and radii respectively equal to B/" and 
B'/'(Fig. 92) describe arcs intersecting in/', and with/' and 
B as centres and radii respectively equal to F A and B/ 
•(Fig. 92) describe arcs intersecting in /. With / and /' as 
centres and radii respectively equal tofh" andf'h' (Fig. 92) 



230 THE TINSMITHS' PATTERN MANUAL. ' 

describe arcs intersecting m A', and with l! and /as centres 
and radii respectively equal to 11 A and /A (Fig. 92) describe 
arcs intersecting in h. Witli h and h' as centres and radii 
respectively equal to 7t g" and h' g' (Fig. 92) describe arcs inter- 
secting in g\ and with g' and Ji as centres and radii respectively 
equal to G A- and hg (Fig. 92) describe arcs intersecting in 
g. VVith g and g' as centres and radii respectively equal to 
g e" and g' e' (Fig. 92) describe arcs intersecting in e', and 
with e' and g as centres and radii respectively to E A and g e 
(Fig. 92) describe arcs intersecting' in e. With e and e' as 
centres and radii respectively equal to e D" and e' D' (Fig 92) 
describe arcs intersecting in D', and similarly with D' and e 
as centres and radii respectively equal to D D" and e D 
(Fig. 92) describe arcs intersecting in D. Join D D'. 
Through B, /, h, g, e, D drav/ an ■ unbroken curved line. 
Also tiiTough B',/', Ji, g\ c', D' draw an unbroken curved 
line. Then BgDD' g' B' is the pattern rec^^uired. 



PROBLEM x::x. 

To draw the patiern fur an ohlong taper hath, the Wee dimensions 
to those fur Problem XIII. being given. 

Again, it is only necessary to treat of two cases — one in this 
problem, and one in the problem foIloAving (see aho § 79, 
p. 157). 

Drav/ (Fig. 94) the plan of the body (sec Problem XIII., 
p. 140), preserving of its construction the centres 0,0'; and 
the points b, b', a, a\ s, s', h, h'. A, A', B, B', in which the 
straight lines and arcs meet each other. Join b, h\ a, a', s, s\ 
Ji,h', A A' (two places), and B B' (two places) as shown in 
the fig. Esamining the plan wo see (rZ, p. 55) that each 
round corner A A' B B' of the toe is the same portion of a 
right cone frustum ; and each of the round corners a a' b' b, 
s s li! a, of the bead are the pame portion {g, p; 129) of an 
oblique cone frubtixm. As we proceed it will be seen that 



THE TINSiilTHS' PATTEKN MANUAL. 



231 



the construction of the pattern for the round ccTners of the 
head of bath is exactly the same as thSft for the round 
corners of the body in Problem XXVI. (see also § 86, 
'p. 206). 

In Plate V. (p. 237) is a representation of an oblong 



Fig. 94. 




taper bath, also of an oblique cone Z, the A portion of 
which corresponds to the A' portion of the body, and the 
development of the former is the development of the latter. , 
Patterns when the body is to be made up of four pieces, i 
We will put the seams to correspond with the lines G h', 
G h\ D D', aad G C. Tke poitteriiis required will be three. 



232 



THE TINSMITH.^' PATTERN MAI^UAL. 



one for the liead of tlie lo&th., one for tlie toe, and oiie for 
the sides. The pattern for the toe can be readily drawn Ly 
Problem XXVII., p. 90. Likewise the pattern for the sides. 
The pattern for the head is drawn as follows. 

Fig. 95. 




Draw separately (Fig. 95) a head-corner portion of 
Fig. 94, say a a' V b, thus. Draw an indefinite line S' d 
(Fig. 95), and with any point (corresponding to O, 
Fig. 94) in it as centre and O a (Fig. 94) as radius describe 
an arc b a. Join 0' (Fig. 94) and produce it to cut arc 



The TINSMITHS' PATTERN MANUAL. 233 

ba ill d; make dh and da (Fig, 95) equal respectively to 
dh and da (Fig, 94). Now (Fig.- 95) make dO' equal to 
dO' (Fig. 94), and with O' as centre and O' a' (Fig. 94) as 
radius describe an arc h' a'. Make d' b' and d' a' equal 
respectively to d'b' and d' a' (Fig 94), Joining bb\ a a' 
completes tlie portion of Fig. 94 required. Now divide 
(Fig. 94) ba into any number of paits. It is convenient 
to t^ke d as one of the division points, and to make d c equal 
to da; leaving cb without further division, thus making 
the division of 6 a into three portions, not all equal. In 
actual practice the dimensions of the work will suggest the 
number of parts necessary. Here 6 c is left without further 
division in order to make clear the correspondence of this 
problem to Problem XXVI., p. 205. Now (Fig. 96) make 
d c equal to d c (Fig. 94), and then b a will be divided 
correspondingly to ba (Fig. 94). Draw XX parallel to 
S' d ; and at d and O draw dD, OQ perpendicular to S' d ; 
and meeting X X in D and Q ; also through d' and 0' draw 
d' D', 0' O" perpendicul ir toS'd; the line O' O" cutting X X 
in Q'. Make Q'O" equal to the given height of the bath, 
and diaw 0"D' parallel to XX, and cutting d' D' in D'. 
Join D D', Q 0" ; produce them to intersect in S (§ 80, 
p. 158) ; and from S let fall a perpendicular to S' d, cutting 
8' d' in S'. "With S' as centre and S' a, or S' c (which is equal 
to S' a) and S' 6 successively as radii describe arcs cutting 
S' d in ^ and /. Draw gf C, /B perpendicular to X X and 
cutting it in C and B ; and join CS, B S. cutting 0''D' in 
0' and B'. 

Next draw SD (Fig. 9Q) equal to S D (Fig. 95) and with 
S as centre and S C, S B (Fig. 95) successively as radii 
describe arcs c and b. With D as centre and radius equal 
to da (Fig. 95) describe arcs cutting arc c in A and C. 
With C as centre and radius cb (Fig. 95) desci'ibe an arc 
cutting arc b in B. Join A, C, and B to S. Make S D' 
equal SD' (Fig. 95) and with S as ci>ntre and S C (Fig. 95 j 
describe an arc (not shown in the fig.) cutting S A and S C 
in A and C respectively ; make S B' equal to S B' (Fig. 95). 



234 



THE TlNSMiTHS' PATTERN MANUAL/ 



Through A, D, C, and B draw an unbroken curved line. 
Also through A', D', C, and B' draw an unbroken curved 
line ; this will complete the pattern for a head-corner. To 
attach the patterns for the flat portions of the head to A A*^ 

, Fig. 96. 




and B B' respectively ; draw through a' (Fig. 94) a line a'F^ 
perpendicular to a! s', and through b' draw 6' G pei-pendicular 
to b' D'. 

Now draw (Fig. 94) a' P perpendicular to a' F and equak 



THE TINSMITHS' PATTERN MANUAL. 235 

to the given height of the bath, asud join F P, then F P is 
the true length of F a'. 

Next draw h' K perpendicular to b' G (h' K will of course 
coincide with the line b' B') and equal to the given height ; 
join G K, then G R is the true length of G 6'. Now with B' 
(Fig. 96) aa centre and G R (Fig. 1)4) as radius, and B as 
centre and radius b G (Fig. 94), describe arcs intersecting 
in G. Join B G, B' G. With A' (Fig. 96) as centre and F P 
{Fig, 9-ij as radius, and A as centre and radius a F (Fig. 94) 
describe arcs intersecting in F. Join A F and produce it 
indefinitely", and make As equal to as (Fig. 94); through 
A' draw A's' parallel to A s and equal to a' s' (Fig. 94); and 
join ss'. The pattern for the portion, seen in plan in Fig. 94, 
Gb ass' a' b' of the head of the bath is now completed. It is 
needless to work out in detail the addition of the portion 
(Fig, 96; sh G h'fs' of the pattern, by which we complete 
the head pattern G E G ^' E' B'. The extra lines ia this 
latter portion of the pattern appertain to the next problem. 



PEOE-LEM XXXI. 

iTo draw, without long radii, the pattern for an oblong taper] 
bath ; given dimensions as in Problem XXX. 

This problem is a second case of the preceding. 

Patterns when the body is to be made up of four pieces- 
' with seams as in preceding problem. 

1 Again, the patterns required will be three ; one for the 
head of the bath, one for the toe, and one for the sides. The 
latter pattern needs no description. The pattern for the toe 
can be readily drawn by Problem XXVIII,, p. 94. The 
pattern for the head can be drawn as follows. 

Draw half the plan of the bath, as the lower half of Fig. 
94, and divide the arcs s h, s' h', each into any number of 
equal parts, here two, in respectively the points/ and/', and 
join//'. Draw (Fig. 94a) any two lines K S, K L perpen- 
dicular to each other, and make K L equal to the given 



236 ' THE TINSMITHS' PATTERN MANUAL. 

heigttt of the batli. From E along K S mark oif K H equal 
to h h",, K F equal to //*, and K S equal to s s' ; and join L H, 
L F, L S ;■ then L H, L F, and L S are the true lengths of 
hH, ff, and s s' respectively. Next join (Fig. 94) /' 7*, s'/; 
draw //", s' s", perpendicular to /' h, s' f respectively, and 
each equal to the given height; and join hf",fs"; then hf", 
fs" may be taken respectively as the true lengths of f h and 
s'/. The true length of h' G may be found along /*' B' as 
was that of 6 G in Problem XXX. along- V B' ; it will of 
course be equal to G R, and we shall speak of it as G R. 

Next draw (see Fig. 96, left-hand portion) ss' equal to 
LS (Fig. 94a), and with s' and s as centres and radii 
respectively equal to f s" and sf (Fig. 94) describe arcs 
intersecting in /. V/ith / and s' as centres and radii 
respectively equal to LF (Fig, 94ffl) and s'/' (Fig. 94) 
describe arcs intersecting in/'; and with/' and / as centres 
and radii respectively equal to hf" SiJxdfh (Fig. 94) descrilio 
arcs intersecting in h ; also with h and /' as centres and 
radii respectively equal to LH (Fig. 94a) and/'/t' (Fig. 94) 
describe arcs intersecting in h'. With h' and ^ as centres 
and radii respectively equal to G E and h G (Fig. 94) 
describe arcs iutersecting in G. Through s,f,Ji, draw an 
unbroken curved line. Also through s',f',h' draw an un- 
broken curved line ; and join li G, G h'. Then shG h' f's' is 
the pattern for the portion of the head of the bath repre- 
Bt-nted in plan in Fig. 94 by the same lettering. It is unne- 
gsssary to pursue the pattern further. 



INDRX. 



The references are to pag-es, except those in brackets, which are to 
paragraphs. 

A. 

« 

Angle, defined, 3; to draw equal to a given angle, 6; to bisect, 10. 

Angles, measurement of, 20. 

Apex, of cone, 24, 105 (43); of pyramid, 66 (29). 

Arc, defined, 5; to complete circle from, g; to find if given curve 

is, 10. 
Arcs, proportionate and similar, 124. 
Athenian hip-bath, plan, 137. 
Axis, of cone, 24 (6), 105 (43); of ellipse, 16; of pyramid, 66 (29). 

B. 

Baking Pan pattern, jj. 

Bath, Athenian hip, plan, 137. 

, 'equal-end,' equal taper pattern, 84-90; four pieces, 84; two 

pieces, 86; one piece, 87; short radius method, 89. 

, — -, unequal taper, plan, 130; pattern, 157-68; four pieces. 

158, two pieces, 161; one piece, 162; when ends are cylindri- 
cal, 164; short-radius method, 166. 

, oblong taper, plan, 140; representation (plate) 237; plate ex- 
plained, 231; pattern, 230-6; short-radius method, 235. 

, Oxford hip, see that heading. 

, oval, plan, 131; representation (plate) 18 1; plate explained, 

169; pattern, 168-83; four pieces, 171; two pieces, 17^; one 
piece, 177; short-radius method, 177. 



240 . INDEX. 

Bath, sitz, plan, 137. 

Bevel (angle). 6. 

Bisect a line, to, 8; an angle, 10. 



c. 

Canister-top, oval, plan, 132; representation (plate), 203; plate 
explained, 193; pattern, 192-205; four pieces, 193; two pieces, 
197; one piece, 198; short-radius method, 200. 

Center of circle, 5; of ellipse, 16. 

Chord, defined, 5. 

Chords, scale of, to draw, 21; how to use, 22. 

Circle, defined, 5; sector of, 28 (9); to find centre, 9; to describe, 
that shall pass through three points, 9; to complete from arc, 
10; to find if given curve is arc of, 10; to inscribe polygon in, 
10; to find length of circumference geometrically, 12. 

, , inclined, extreme cases of oblique cone frusta, 112 (61); 

pattern for, 121, 

Circumference of circle defined, 5; to find length of, geometrically, 
12. 

Classification of patterns, 2. 

Coffee-pots, 34; hexagonal, 70 (35). 

Cone, defined, 105 (43); axis, radius, apex, base, 24 (6), 105 (43); 
elevation of generatmg lines, 108 (55). 

, right, defined, 24 (6), 105 (44); basis of patterns for articles of 

equal taper, 24 (5); compared with oblique case, 106 (from 47); 
representation (plate), 227; development of, by paint, 28; gen- 
erating lines, 106(47); corresponding points of generating lines, 
see Corresponding points; to find height, 26; to find slant, 27; 
to find dimensions of, from frustum, 36. 

Cone, right, pattern, 29-31 ; one piece, 29; more than one piece, 30. 

, , frustum (round equal-rapering body), defined, 33; rep- 
resentation, 50; representations of round equal-tapering arti- 
cles, 34; relations of, with complete cone, 34 (from 13); de- 
velopment, 34 (15). 

-, , , plan, 50 (23); characteristic features, 55 {c, d)\ to 

draw, see equal-tapering bcdies (plans, to draw) 

, , , pattern, 37-43; ends and height given, 47; ends and 

slant given, 39; pattern for parts of, 39; short-radius method, 41. 

-, oblique; defined, 106 (45); basis of patterns for articles of un- 
equal taper, 105 (42); compared with right cone, 106 (from 47); 



INDEX. 241 

obliquity, how measured, 107 (52); representations (plates), i8r, 
203, 213, 227, 237; development of, 108 (56); generating lines, 
106 (46-7); longest and shortest generating lines, 107 (from 51); 
lines of greatest and least inclination, 107 {c,2); height of ele- 
vations of generating lines, 108 (55); true lengths of elevations 
of generating lines, 107 (54). 

, , plan of axis, 125-6 (a); of generating lines, 126 (<r/, e); of 

longest and shortest generating lines, 126 (d); of apex, 126 
(from c). 

, , pattern, 108. 

, , frustum (round unequal-tapering body), defined, in (58- 

9); representation, 125 (plate); generating lines, 126 {d); 
generating lines of, when circumscribing oblique pyramid 
frustum, 150 (78); corresponding points of generating lines, see 
corresponding points; oblique cylinder, an extreme case of, 
112 (61). 

, , , plan, 125 (69), 128 (from 73); of axis, 126 (a); of 

lines of greatest and least inclination (longest and shortest 
generating lines), 126 {d); characteristic features, 127 (71-73). 

, , , pattern, 113-23; ends, height, and inclination of 

longest generating line given, 113; height and plan given, ti6; 
short-radius method, apex accessible or inaccessible, 118; ex- 
treme case of (oblique cylinder), 121. 

Corresponding points, of right cone frustum, 35,51; of equal-ta- 
pering body, 52 (25); of oblique cone frustum, 124 (68), 126 (d). 

, in plan, of right cone frustum, 51 (23, 24) 54; of equal- 
tapering body, 52 (25), 53 (26); of oblique cone frustum, 124 (68), 
129 (^); of oblique pyramid frustum, 145 {76); of oval equal-ta- 
pering body, 65 (28); distance be'tvveen, is ec[ual in plans of 
equal-tapering bodies, 55 (from a); to find distance between, 
height and slant given, 56; height and inclination given, 57. 

Cylinder, right, defined, 112 (61). 

, oblique, defined, 112 (61); an extreme case of oblique cone 

frustum, 112 (61); pattern for, 121. 

D. 

Definitions, of straight lines, angle, perpendicular, 3; parallel 
lines, triangle, hypotenuse, polygon, pentagon, hexagon, hep- 
tagon, octagon, quadrilateral, square, oblong, rectangle, 4! 
circle, circumference, arc, quadrant, semicircle, radius chord, 



342 INDEX. 

diameter, 5; ellipse, 17; ellipse focus, axis, diameter, centre, 
16; cone, 105 (43); cone axis, radius, apex, base, 24 (6), 105 (43) 
right cone, 24 (6), 105 (44); right cone frustum, 33; oblique 
cone, 106(45); oblique cone frustum, in (58-9); cylinder, right 
and oblique, 112 (61); pyramid and axis, 66; pyramid frustum, 
143(74); proportional arcs, similar arcs, 124; corresponding 
points, see Corresponding points. 

Degree (angle), explained, 21, 

Diameter of circle, 5. 

of ellipse, 16. 



E. 



Edging, allowance for, 33. 

Egg-shaped oval, to draw, 14. 

Elevation, explained, 48. 

Ellipse, defined, 17; focus, axes, diameter, centre, 16; to describe 

mechanically, 15, 18; geometrically, 17. 
Equal end bath, see Bath. 
Equal-tapering bodies, 24. 
Round, 28-45; essentially right cone frusta, 34 (13); to find slant, 
or height of cone of which body is portion, 36; to find slant of 
body, ends and height given, 43; to find height, ends and slant 
given, 44; slant and inchnation given, 44; see also Cone (right, 
frustum). 
Of flat surfaces, 66-83; essentially right pyramid frusta, see Pyra- 
mid (right, frustum). 
Of fiat and curved surfaces, 84-96; curved surfaces, portions of 
right cone frusta, 55 [d), 84; see also Cone (right, frustum). 

, plans. 

Characteristic features of, of round bodies, 51; characteristic 
features, body oblong with round corners, 52; features of plans 
summarized, 55; how to find from plan if article is of equal 
taper, 55 {b). 

, , to draw. 

Either end, height, and slant given, 57; either end and distance 

between corresponding points ('out of flue') given, 59. 
Oval bodies, 64. 

Of flat and curved surfaces. Oblong bodies, 59, 62. 
See also Corresponding points. 



INDEX. 243 

, patterns. 

Round bodies (frusta of right cones), 28-45; see also Cone (right, 

frustum, pattern). 
Oval bodies. Patterns, 96-104; in four pieces, 96; in two pieces, 
99; in one piece, 100; short radius method, 102. 

. Bodies having flat surfaces. 

Ends of body and height given, 71; short radius method, 73; body 
oblong or square (pan) pattern in one piece with bottom, "]"], 80 
same with bottom, sides, and ends in separate pieces, 82 
baking-pan pattern, bottom, width of top, and slant given, "]"] 
bottom, length of top, and slant given, 80; top, slant, and height 
given, 81; top, slant and inclination of slant given, 8[. 

• . Bodies of flat and curved surfaces combined. 

Body having flat sides and semicircular ends, see Bath (equal 
end and equal-taper); flat sides and ends, and round corners 
(oblong or square with round corners), see Oblong pan. 

F. . 

Flue, out of, 55, 61. 
Focus of ellipse, 16. 

Frustum of cone, see Cone (right, frustum; and oblique, frustum); 
of pyramid, see Pyramid (right, frustum; and oblique, frustum). 



Generation of Cone, 24. 

Gravy strainers, frusta of right cones, 34 ^13). 

Grooved seam, 32, 

H. 

Heptagon. 4. 

Hexagon, 4. 

Hexagonal pyramid, 66 (29); pattern of right, see Pyramid (right 

frustum, pattern); oblique, see Pyramid (oblique, frustum 

pattern); coffee-pots, 70 (35). 
Hip baths, see Athenian hip bath, Oxford hip bath. 
Hoods, their relation to truncated pyramids, 70, 143; pattern, 154. 
Hoppers, 143; see also Hoods. 
Hypotenuse, 4. 



244 INDEX. 

I. 

Inclination of Slant, defined, 24(4), 55 [d), 61 (Case II.); articles 
of equal, see Equal-tapering bodies; of unequal, see Unequal- 
tapering bodies. 

Introductory Problems, 3. 

L. 

Lap, allowance for, 32. 

Line, to divide into equal parts, 7; into two parts, 8. 
Lines, straight, defined, 3; points joined by, 7; parallel defined, 4; 
true lengths of, 48. 

o. 

Oblique Cone, see Cone oblique; cylinder, see Cylinder. 

Oblong defined, 4; with round corners, to draw, iq; with semicircu- 
lar ends, to draw, 20; oblong equal-tapering body, see Equal 
tapering bodies. 

— —or square pan with round corners, pattern qo; in four pieces, 

go; in two pieces, 92; in one piece, 93; short-radius method, 94. 

taper bath, see Bath (oblong taper). 

Octagon, 4 

Oval, to draw, 13; egg-shaped, to draw, 14. 

bath, see Bath. 

Canister top, representation (plate), 203; plate explained, 193, 

plan, 132; pattern, 192; in four pieces, 193; in two pieces, 197; 

in one piece, 198. 
equal-tapering bodies, see Equal tapering bodies (plans, to 

draw; and patterns); corresponding points (in plan). 
unequal-tapering bodies, see Unequal-tapering bodies, and 

Unequal-tapering bodies (plans; patterns). 
Oxford hip-bath, representation (plate), 227; plate explained, 217; 

plan, 134; pattern, 216; short radius method, 223. 



Pails, frusta of right cones, 34 (13"). 

Pan, baking, see Baking-pan; oblong with round corners, see ob- 
long pan. 



INDEX. 245 

Parallel lines, 4. 

Pattern, setting out, is development of surfaces, 2 (3\ 34 (14). 
Pentagon, 4. 
Perpendicular, 3. 

Pipes, tapering, are frusta of oblique cones, 112 (5q). 
Plan, explained, 46; of equal-tapering bodies, see Equal-tapering 
bodies (plans); of unequal-tapering bodies, see Unequal-taper- 
ing bodies, (plans). 
Plates, 181, 203, 213, 227, 237; descriptions respectively, 169, 193, 

206, 217, 231. 
Points, angular, 3; always joined by straight lines, 7; corresponding, 

see Corresponding points. 
Polygon, defined, 4; to inscribe in circle, 10; regular, to describe, 11. 
Projection, explained, 47. 
Proportional ^rcs, 124. 
Pyramid, 
Defined, 66 (29); base apex, axis, 66 (29); triangular, square, hex- 
agonal, 66(29). 

, right, 

Defined, 66 (29); can be inscribed in right cone, 66, (from 30); 
model of, 73 (37). 

, — — , pattern, 

Hexagonal, 68; any number of faces, 69. 

, , frustum. 

Defined, 69, (33), 70 (34); representation, 70; can be inscribed in 
right cone frustum, 70 (36); the basis of articles of equal-taper 
having flat surfaces, 70 (35); model, 73 (37); slant of face of, 
151, (Case II). 

, , , pattern. 

Hexagonal, 71; any number of faces, jy, short-radius method, 73. 

, oblique. 

Defined, [43 (74); can be inscribed in oblique cone, 144 (75). 

^ ^ pattern. 

Plan and height given, 145; plan and axis given, 148. 

, , frustum, (unequal-tapering body). 

Defined, 143 (74); the basis of numerous articles of unequal- 
taper having flat surfaces, 143; can be inscribed in oblique 
cone frustum, 150 (78); slant of face of, 151 ; madel, 73 (37). 

, , , plan, 

What it consists of, 144 (.j6)\ corresponding points in, 145 (76) 



240 INDEX. 

how to determine from plan of unequal-tapering body if body 
is oblique pyramid frustum, 145 {77). 

, , , pattern, 

Plan and height given, (frustum, hexagonal), 148; ends and slant 
and inclination of one face given, 151; short radius method 
152. 

Q 

Quadrant, defined, 5; to divide, 11. 
Quadrilaterals, 4. 

Radius, defined, 5; of cone, 24. 
Rake (angle), 6. 
Rectangle, defined, 4. 
Right angle, 3. 

, cone, see Cone (right); pyramid, see Pyramid (right). 

Round articles of equal-taper, see Cone (right frustum); of un- 
equal-taper, see Cone (oblique frustum). 

s. 

Seams, 32. 

Sector, 28. 

Semicircle, 5; to divide, 11. 

Short radius methods — for round equal-tapering body (frustum of 
right cone), 41 ; equal-tapering body of flat surfaces (frustum 
of right pyramid), 72, 75; oblong body of equal-taper and 
semi-circular ends, 89; oblong (or square) body with round 
corners, 94; oval equal-tapering body, 102; round unequal- 
tapering body (frustum of oblique cone), 118; frustum of ob- 
lique pyramid, 152; oblong body of unequal taper and semi- 
circular ends, 166; oval unequal tapering body, 177; body of 
oblong bottom with semicircular ends, and circular top, 189; 
body of oval bottom and circular top, 200; body oblong bot- 
tom with round corners, and circular top, 212; Oxford hip-bath, 
223; oblong taper-bath, 235. 

Similar arcs, 124. 

Sitz-bath, plan, 137. 



INDEX. 247 

Slant, a length not an angle, 24 (4); further defined, 52 (25); efface- 
of frustum of pyramid, 15 [ (Case II.); inclination of, see In- 
clination slant. 

Square, defined, 4; pyramid, 66 (2gj, 

bodies, of equal-taper, essentially right pyramid frusta, see- 
Pyramid (right frustum); also equal-tapering bodies (pat- 
terns), and oblong pan. 

Subtend explained, 20. 

Taper, an angle, 24 (4). 

bath, oblong, see Bath (oblong taper). 

Tapering articles; equal, see Equal-tapering articles; unequal, see- 

Unequal-tapering articles. 
Tea-bottle top pittern, 183-igi; in four pieces, 183; in two pieces 

187; in one piece, 188; short radius method, i-g. 
Triangle, 4. 

Triangular pyramid, 66 (29). 
Truncated pyramid, 143 (74). 

u. 

Unequal-Tapering Bodies. 

Round, essentially oblique cone frusta, 105 (42); see also Cone 
(oblique, frustum). 

Of curved surfaces. Made up mostly of portions of oblique 
cone frusta, as oval bath, 169; body of oval bottom and circu- 
lar top, 193; Oxford hip-bath, 217 . 

Of flat surfaces. Essentially oblique pyramid frusta, see Pyra- 
mid (oblique, frustum). 

Of flat and curved surfaces, 157; the curved surface portions of 
frusta of oblique cones, 157; ia "equal-end" bath, 157; in body 
of oblong bottom with round corners and circular top, 206; in 
body of oblong bottom, with semicircular ends, and circular 
top, 193; in oblong taper bath, 230; see also Cone (ob lique 
frustum). 
, plans. 

How to find from plan if article is of unequal-taper, I2g. 

Round; see Cone (oblique, frustum, plans). 

Of curved surfaces, body oval top and bottom, see Bath (oval); 



248 INDEX. 

bottom oval, top circular, see Oval canister-top; and see also 
Bath (Oxford-hip; Athenian; and Sitz). 

Of flat surfaces. See Pyramid (oblique, frustum, plan). 

Of flat and curved surfaces. Body oblong with semicircular 
ends, see Bath (equal end and unequal-taper); oblong bottom 
with semicircular ends, and circular top, see Tea-bottle top 
body oblong (or square) with round corners and circular top 
133; see also Oblong taper bath. 
-, patterns. 



Round bodies, see Cone (oblique, frustum, pattern). 

Of curved surfaces. Body oval, top and bottom, see Bath (oval) ; 
bottom oval, top circular, 200 (see also Oval canister-top); Ox-' 
ford hip-bath, see Oxford hip-bath. 

Of flat surfaces, see Pyramid (oblique, frustum, pattern). 

Of flat and curved surfaces. Body oblong with semicircular 
ends, see Bath (equal end and unequal-taper); oblong bottom 
with semicircular ends and circular top, i8g (see also Tea- 
bottle top); body oblong (or square) with round corners, and 
circular top, 205; in four pieces, 206; in two pieces, 209; in one 
piece, 210; short-radius method, 212; oblong taper-bath, see 
Bath (oblong taper). 



w. 



Wiring, allowance for, 32. 



THE rOX rURNflCE 




*'1T»S A. MUi^xa-iOii." 



I' 



For Hard Coal 
Soil €oal 
FUKVACES -! " Coke 
I " \* ood 
I " Gas 



IT /S THE FURNACE YOU OUGHT TO SELL. 

FOX FURNACE CO., Cleveland, 0. 



Hot Air Furnace Work 
PAYS, 

IF you sell the right kind of a furnace — one 
that invariably works well, saves fuel and is easily 
managed. Customers are willing to pay a dealer a fair 
price for such a furnace. How comfortable it is for a 
dealer to feel sure that the goods he sells are^sure^ 

to please! 



P 
A 
R 
A 
G 
O 

N 

• 

F 
U 
R 
N 
A 
C 
E 
S 




P 
A 
R 
A 
G 
O 
N 
^ 

F 
U 
R 
IV 
A 
C 
E 
S 



Inspire that Comfortable Feeling: 

THBY ALWAYS SATISFY. 

Send for our furnace book "HINTS ABOUT HEATING." 



Manufacturers. 
1801 N. Fourtli StieeS - - PHIt. i DEL.PHIA., PA. 



Selling: Kitchen Ranges 
PAYS, 

IF you sell ranges that have some specially 
attractive or unique features— something a 
little different from the common run of kitchen goods. 
AVhat do you say about handling just such a range? 
Boiler elevated and out of the way. "Eclipse Covers," 
High Hot Closets. All modern improvements. Will 
make hot water as fast as a horse can trot. Thousands 
an use. 



F 
I 
D 

E 
L 
I 

T 
Y 



R 
A 
N 
G 

E 
S 




F 

I 

D 

E 

L 

I 

T 

Y 



R 
A 
N 
G 

E 
S 



Will ''Put Money in Your PurseJ' 
EVERYBODY LIKES THEM. 

Send to us for circulars and prices. Exclusive territory given. 

ISAAC A. SHEPPARD & CO. 

MANUFACTURERS, 

1801 N. Fourth St., = = Philadelphia, Pa. 



BooKs by /r\ail 



ilit4iii>fe>iiti. 



m-^ . . . 



We will take pleasure in supplying Books of whatever 
character at catalogue prices on receipt of price. 




eooKS ors 

Sheet Metal WopKing, 
Tne Foundry, 
Ttie WoFKsHop, 
Heating and Ventilation, 
Meetianieal Dr'awing, 
Plumbing and DFainage, 
Ttie Store and Office, 

AFf~ ESPECIALLY IM OUR LINE. 




^ 



J\DDRESS 



STEBN, PnbMer aod Bocheller, 

>9 Dearborn St., CHICAGO, 










s 1 liiLUl 












B 






Manufacturers of. 



Plain, Retinned, Galvanized, Japanned 
and Enameled, Pieced and Stamped 

TINWARE 



AND SHEET-IRON GOODS 

St. Paul Ave., Ninth and Tenth Sts,, 
MILWAUKEE, WIS. 



...SEND FOB CATALOGUE AND PRICES... 



CHICAGO BRANCH:; 

78 and 80 Lake Street. 




SRetal Starnping 5: Spinning Gq. 

GRAND RAPIDS, HICH. 
*^^y^ Factory «& Office, 423-429 Straight St. 

. . MANUFACTURERS OF . . 

Arehiteetural Sheet Metal Ornaments for 
Inside and Outside Decorations, 

Of Zinc, Copper, Brass and Alaminnm. 

Steel Ceilings. 
\ Pieced, Stamped and 




€1 



i) 



Japanned, and non- 
rusting Tinivare. 

Metal Spinning in 
Brass, Copper, Zinc 
and Alaminnm. 

Model Making in 

l¥ood an<l Metal. 

Estimates on work cheerfully 
given. 




A Handy Book 
for Tinners 



TINSMITHS' 
AND SHEET METAL WORKERS' 
POCKET REFERENCE BOOK. 

A collection of practical infofmation, 
including rules, tables, receipts, explana- 
tions, etc., used' daily by the tinsmith at his 
work. Presented in a compact form, con- 
venient for carrying in the pocket. 

By C. E. BOD LEY. 



•^ A e vLdVI/— a A ^ 



Price, postage prepaid, fine manilla paper covers, 
35'c.; clotli, 5'oc. 



"^^ A to \I^S/ -a~A '"? 



DANIEL STERN, 

69 DEARBORN STREET, ChICAGO. 



Bmeilcaii Tin aim Teine Rales. 



/^ 



HAMILTON'S 

BEST RE'DIRRED ^f^ 



The very best Roofing Plate ever made. 
Perfectly re=squared and guaranteed everlasting. 

Osceola Old Style 

mm OLD PROCESS... BONUS... :: :: 

A good old style plate, A fine extra covered plate, 

at a reasonable price. and a great favorite. 

Purely a hand-coated, redipped plate, and guaranteed equal to 
any ''Extra Coated" plate on the market. 

LULU-^Qood ordinary plate. KILLBUCK— riedium Grade 

AH the Above Brands in Stock. 



tJohn Hamilton, 

Tin Plate Manufactarer, 

62-64 THIRD AVE. PITTSBURGH, PA. 




LARGEST LINE OF. 



Portable Ranges, Cylindev Stoves, 
Parlor Stoves, liaiiiidry Stoves 
AND Furnaces in the world... 



UNION STOY£ WORKS, 70 Beekman Street, N. Y. 



lOO 

TINNERS' 





^l.iF 



The American Artisan Full Size Patterns, 
printed on manilla paper, from which they are 
readily transferred to heavy sheets and cut out 
ready for use. Price, sent post-paid, for the 

Full Set of 100 Patterns, 

l.OO. 

ADD-RESS 

DANIEL STERN, 

PUBLISHER, 

69 Dearborn Street, Chicago, III. 



F? RLaIABLaE ^' ii * 




AND 
FOR 

HRATINQ 

AND 

COOKINQ 
ARE THE BEST„ 



too STYLES AND SIZES. 



,tt?2Ei!\ 



SEND FOR 
GATAIaOGUE. 



's^P' 




iehnei^ep ^ (^penkamp 



CLEVELAND, OHIO. 



LIGHTNING SLATE DRESSER. 




FOR SLaAXR ROOFRRS. 

Will do the work of ten to twenty men and do it 
much better than by hand; adjustable to any size or 
shape; cuts and punches at same operation; counter- 
sinks the hole;- cuts round or straight; spoils no slate. 

PATENTED AND MANUFACTURED BY 

CLEVELAND, OHIO. 

Manufacturers and Dealers in all kinds of 

Roofing Slate, Slate Roofers' Supplies, Slaters' 
Tools and Slate Blackboards. 

Publishers of the Slate Roofers' Book of Instructions; 
by mail, gi.oo. 

CORRESPONDENCE SOLICITED. 



BOOK J PT nmi. 



We will take pleasure in supplying Books of what- 
ever character at catalogue prices, prepaid by 
mail, to any address, on receipt of price. 



Sheet Metal Worker's Guide. 

By W. J. E. Crane. Price, postpaid, 60 cents. 

Tin, Sheet Iron and Copper Plate Worker. 

By Leroy J. Blinn. With over 100 illustrations, $2.50. 

Galvanized Iron Cornice-Worker's Manual. 

By Charles A. Vaile. Illustrated by twenty-one plates. In 
one volume, 4to, $5.00. 

The Sheet-Metal Worker's Instructor. 

By Reuben H, Warn. $3.00. 

Tinsmith's Helper and Pattern Book. 

By H. K. Vosburgh. Price, $1.00. 



In addition to the above, a number of Standard 
Works on Heating and Ventilation, Plumbing and 
Draining, and the Store and Ofifice are kept in stock, 
and will be supplied at publishers' prices. 

DANIEL STERN, 

Publisher and Bookseller, 
69 Dearborn Street, - - CHICAGO, ILL. 



THE BOYNTON FURNACE CO., 



Sole Manufacturers of 



BOYNTON'S FDRNACES, RANGES, 

Hot Water and Steam Heaters. 



V 



■« 



■Ml 

I 



0? 







tq 






207, 209 and 211 Water St., New York. 
195 and 197 Lake St., Chicago. 

N. A. BoYNTON, Pres ) 

C. B. BoYNTON, Vice-Pres. [ lYElV YORK^ 

E. E, Dickinson, Sec. and Treas. ) 

J.H.Manny, ? ,- ^ltt^^^/-^ 

C. E. Manny, 5 Managers, CHICAGO. 

Send for Catalogues an«l Discounts-* 



f IS 






